Scalar Triple Product MCQ Quiz in हिन्दी - Objective Question with Answer for Scalar Triple Product - मुफ्त [PDF] डाउनलोड करें

अवधारणा:

1. वे सदिश जो एक ही समतल के समानांतर हैं या एक ही समतल पर स्थित हैं, उन्हें समतलीय सदिश कहा जाता है।

2. तीन सदिश समतलीय होते हैं यदि उनका अदिश त्रिक गुणनफल शून्य होता है। $⃗a\cdot (⃗b\times ⃗c)=0$

गणना:

मान लीजिए $\overrightarrow{a}$ = 2i - j + k

$\overrightarrow{b}$ =i + 2j - 3k

$\overrightarrow{c}$= 3i + a j + 5k

$⃗b\times ⃗c=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& -3\\ 3& a& 5\end{array}\right|$

⇒ $\hat{i}(10+3a)-\hat{j}(5+9)+\hat{k}(a-6)$

⇒ $(10+3a)\hat{i}-14\hat{j}+(a-6)\hat{k}$

⇒ $\overrightarrow{a}\cdot (\overrightarrow{b}\times \overrightarrow{c})=(2\hat{i}-\hat{j}+\hat{k})\cdot ((10+3a)\hat{i}-14\hat{j}+(a-6)\hat{k})$

⇒ 2(10 + 3a) - (-14) + (a - 6)

⇒ 20 + 6a + 14 + a - 6

⇒ 7a + 28

चूँकि 7a + 28 = 0

⇒ a = -4

अतः विकल्प (4) सही है।

संकल्पना:

तीन सदिशों $\overrightarrow{A},\overrightarrow{B}$ और $\overrightarrow{C}$ के अदिश त्रिक गुणनफल को निम्न द्वारा निरूपित किया जाता है,

• $[\overrightarrow{\mathrm{A}},\overrightarrow{\mathrm{B}},\overrightarrow{\mathrm{C}}]=\overrightarrow{\mathrm{A}}\cdot (\overrightarrow{\mathrm{B}}\times \overrightarrow{\mathrm{C}})=\overrightarrow{\mathrm{B}}\cdot (\overrightarrow{\mathrm{C}}\times \overrightarrow{\mathrm{A}})=\overrightarrow{\mathrm{C}}\cdot (\overrightarrow{\mathrm{A}}\times \overrightarrow{\mathrm{B}})$
• $[\overrightarrow{\mathrm{A}},\overrightarrow{\mathrm{B}},\overrightarrow{\mathrm{C}}]=\left|\begin{array}{ccc}{A}_1& A_2& A_3\\ B_1& B_2& B_3\\ C_1& C_2& C_3\end{array}\right|$

जहाँ, $\overrightarrow{\mathrm{A}}={\mathrm{A}}_1\hat{\mathrm{i}}+{\mathrm{A}}_2\hat{\mathrm{j}}+{\mathrm{A}}_3\hat{\mathrm{k}}$, $\overrightarrow{\mathrm{B}}={\mathrm{B}}_1\hat{\mathrm{i}}+{\mathrm{B}}_2\hat{\mathrm{j}}+{\mathrm{B}}_3\hat{\mathrm{k}}$ और $\overrightarrow{\mathrm{C}}={\mathrm{C}}_1\hat{\mathrm{i}}+{\mathrm{C}}_2\hat{\mathrm{j}}+{\mathrm{C}}_3\hat{\mathrm{k}}$

गणना​:

हमें तीन गैर समतलीय सदिश $\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}}$ और $\overrightarrow{\mathrm{c}}$ प्राप्त हैं, 

∴ $[\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}}]=\overrightarrow{\mathrm{a}}\cdot (\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}})=\overrightarrow{\mathrm{b}}\cdot (\overrightarrow{\mathrm{c}}\times \overrightarrow{\mathrm{a}})=\overrightarrow{\mathrm{c}}\cdot (\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}})$

∴ $\frac{\overrightarrow{\mathrm{a}}.(\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}})}{\overrightarrow{\mathrm{b}}.(\overrightarrow{\mathrm{c}}\times \overrightarrow{\mathrm{a}})}+\frac{\overrightarrow{\mathrm{b}}.(\overrightarrow{\mathrm{c}}\times \overrightarrow{\mathrm{a}})}{\overrightarrow{\mathrm{c}}.(\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}})}+\frac{\overrightarrow{\mathrm{c}}.(\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{a}})}{\overrightarrow{\mathrm{a}}.(\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}})}$

= $\frac{[\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}}]}{[\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}},\overrightarrow{\mathrm{a}}]}+\frac{[\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}},\overrightarrow{\mathrm{a}}]}{[\overrightarrow{\mathrm{c}},\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}}]}+\frac{[\overrightarrow{\mathrm{c}},\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{a}}]}{[\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}}]}$

= $\frac{[\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}}]}{[\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}}]}+\frac{[\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}}]}{[\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}}]}-\frac{[\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}}]}{[\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}}]}$

= 1 + 1 - 1

= 1

∴  $\frac{\overrightarrow{\mathrm{a}}.(\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}})}{\overrightarrow{\mathrm{b}}.(\overrightarrow{\mathrm{c}}\times \overrightarrow{\mathrm{a}})}+\frac{\overrightarrow{\mathrm{b}}.(\overrightarrow{\mathrm{c}}\times \overrightarrow{\mathrm{a}})}{\overrightarrow{\mathrm{c}}.(\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}})}+\frac{\overrightarrow{\mathrm{c}}.(\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{a}})}{\overrightarrow{\mathrm{a}}.(\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}})}$ का मान 1 है। 

सही उत्तर विकल्प 3 है।

संकल्पना:

• $[\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}}]=\overrightarrow{\mathrm{a}}\cdot (\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}})=\overrightarrow{\mathrm{b}}\cdot (\overrightarrow{\mathrm{c}}\times \overrightarrow{\mathrm{a}})=\overrightarrow{\mathrm{c}}\cdot (\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}})$
• $\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}=|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}|\sin \theta$
• $\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}}=|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}|\cos \theta$

गणना​:

दिया गया है, $\overrightarrow{\mathrm{c}}$, $\overrightarrow{\mathrm{a}}$ और $\overrightarrow{\mathrm{b}}$ दोनों के लंबवत एक इकाई सदिश है।

∴ $\overrightarrow{\mathrm{c}}$ = λ $(\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}})$

$|\overrightarrow{\mathrm{c}}|=1$

हम जानते हैं कि $[\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}}]=\overrightarrow{\mathrm{a}}\cdot (\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}})=\overrightarrow{\mathrm{b}}\cdot (\overrightarrow{\mathrm{c}}\times \overrightarrow{\mathrm{a}})=\overrightarrow{\mathrm{c}}\cdot (\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}})$

${[\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}}]}^2={[\overrightarrow{\mathrm{c}}\cdot (\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}})]}^2$

दिया गया है, $\overrightarrow{\mathrm{c}}$, $\overrightarrow{\mathrm{a}}$ और $\overrightarrow{\mathrm{b}}$ दोनों के लंबवत एक इकाई सदिश है।

∴   $\overrightarrow{\mathrm{c}}$, $\overrightarrow{\mathrm{a}}$ × $\overrightarrow{\mathrm{b}}$ के समानांतर है।

∴ हम समीकरण को इस प्रकार लिख सकते हैं:

${[\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}}]}^2=||\overrightarrow{\mathrm{c}}||(\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}})|\cos 0{|}^2$

⇒ ${[\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}}]}^2={||\overrightarrow{\mathrm{c}}||(\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}})||}^2$

⇒ ${[\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}}]}^2={||\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}|\sin \frac{\pi }{6}|}^2$

⇒ ${[\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}}]}^2={\left|\frac{|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}|}{2}\right|}^2$

⇒ ${[\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}}]}^2=\frac{|\overrightarrow{\mathrm{a}}{|}^2|\overrightarrow{\mathrm{b}}{|}^2}{4}$

अभीष्ट मान ${[\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}}]}^2=\frac{|\overrightarrow{\mathrm{a}}{|}^2|\overrightarrow{\mathrm{b}}{|}^2}{4}$ है। 

संकल्पना:

तीन सदिशों $\overrightarrow{A},\overrightarrow{B}$ और $\overrightarrow{C}$ के अदिश त्रिक गुणनफल को निम्न द्वारा निरूपित किया जाता है,

• $[\overrightarrow{\mathrm{A}},\overrightarrow{\mathrm{B}},\overrightarrow{\mathrm{C}}]=\overrightarrow{\mathrm{A}}\cdot (\overrightarrow{\mathrm{B}}\times \overrightarrow{\mathrm{C}})=\overrightarrow{\mathrm{B}}\cdot (\overrightarrow{\mathrm{C}}\times \overrightarrow{\mathrm{A}})=\overrightarrow{\mathrm{C}}\cdot (\overrightarrow{\mathrm{A}}\times \overrightarrow{\mathrm{B}})$
• $[\overrightarrow{\mathrm{A}},\overrightarrow{\mathrm{B}},\overrightarrow{\mathrm{C}}]=\left|\begin{array}{ccc}{A}_1& A_2& A_3\\ B_1& B_2& B_3\\ C_1& C_2& C_3\end{array}\right|$

जहाँ, $\overrightarrow{\mathrm{A}}={\mathrm{A}}_1\hat{\mathrm{i}}+{\mathrm{A}}_2\hat{\mathrm{j}}+{\mathrm{A}}_3\hat{\mathrm{k}}$, $\overrightarrow{\mathrm{B}}={\mathrm{B}}_1\hat{\mathrm{i}}+{\mathrm{B}}_2\hat{\mathrm{j}}+{\mathrm{B}}_3\hat{\mathrm{k}}$ और $\overrightarrow{\mathrm{C}}={\mathrm{C}}_1\hat{\mathrm{i}}+{\mathrm{C}}_2\hat{\mathrm{j}}+{\mathrm{C}}_3\hat{\mathrm{k}}$

गणना​:

हमें तीन गैर समतलीय सदिश $\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}}$ और $\overrightarrow{\mathrm{c}}$ प्राप्त हैं, 

∴ $[\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}}]=\overrightarrow{\mathrm{a}}\cdot (\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}})=\overrightarrow{\mathrm{b}}\cdot (\overrightarrow{\mathrm{c}}\times \overrightarrow{\mathrm{a}})=\overrightarrow{\mathrm{c}}\cdot (\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}})$

∴ $\frac{\overrightarrow{\mathrm{a}}.(\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}})}{\overrightarrow{\mathrm{b}}.(\overrightarrow{\mathrm{c}}\times \overrightarrow{\mathrm{a}})}+\frac{\overrightarrow{\mathrm{b}}.(\overrightarrow{\mathrm{c}}\times \overrightarrow{\mathrm{a}})}{\overrightarrow{\mathrm{c}}.(\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}})}+\frac{\overrightarrow{\mathrm{c}}.(\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{a}})}{\overrightarrow{\mathrm{a}}.(\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}})}$

= $\frac{[\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}}]}{[\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}},\overrightarrow{\mathrm{a}}]}+\frac{[\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}},\overrightarrow{\mathrm{a}}]}{[\overrightarrow{\mathrm{c}},\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}}]}+\frac{[\overrightarrow{\mathrm{c}},\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{a}}]}{[\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}}]}$

= $\frac{[\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}}]}{[\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}}]}+\frac{[\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}}]}{[\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}}]}-\frac{[\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}}]}{[\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}}]}$

= 1 + 1 - 1

= 1

∴  $\frac{\overrightarrow{\mathrm{a}}.(\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}})}{\overrightarrow{\mathrm{b}}.(\overrightarrow{\mathrm{c}}\times \overrightarrow{\mathrm{a}})}+\frac{\overrightarrow{\mathrm{b}}.(\overrightarrow{\mathrm{c}}\times \overrightarrow{\mathrm{a}})}{\overrightarrow{\mathrm{c}}.(\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}})}+\frac{\overrightarrow{\mathrm{c}}.(\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{a}})}{\overrightarrow{\mathrm{a}}.(\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}})}$ का मान 1 है। 

सही उत्तर विकल्प 3 है।

संकल्पना:

1) तीन सदिशों के समतलीय होने के लिए, उनका अदिश त्रिक गुणनफल शून्य होना चाहिए।

2) सदिशों के अदिश त्रिक गुणनफल $\overrightarrow{a},\overrightarrow{b}$ और $\overrightarrow{c}$, $[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}]$ द्वारा निरूपित किया जाता है।

गणना:

दिए गए सदिश λî + ĵ + 2k̂, î + λĵ - k̂ और 2î - ĵ + λk̂ हैं।

तीन सदिशों के समतलीय होने के लिए, उनका अदिश त्रिक गुणनफल शून्य होता है।

⇒ $\left|\begin{array}{ccc}\lambda & 1& 2\\ 1& \lambda & -1\\ 2& -1& \lambda \end{array}\right|$ = 0

⇒ λ(λ² - 1) - (λ + 2) + 2(- 1 - 2λ) = 0

⇒ λ³ - 6λ - 4 = 0

⇒ (λ + 2)(λ² - 2λ - 2) = 0

हल करने पर हम प्राप्त करते है; λ = - 2 या λ = $1\pm \sqrt{3}$

इसलिए, सदिश λ = - 2 के लिए समतलीय हैं।

सही उत्तर विकल्प 1 है।

अवधारणा:

अदिश त्रिक गुणनफल के गुण

• [a b c] = [b c a] = [c a b]
• [a b c] = - [b a c] = - [c b a] = - [a c b]
• [(a + b) c d] = [a c d] + [b c d]
• [λa, b c] = λ [a b c]
• तीन गैर-शून्य सदिश $\overrightarrow{a},\overrightarrow{b}और\overrightarrow{c}$ समतलीय हैं यदि और केवल यदि [abc] = 0

गणना :

दिया गया: $\overrightarrow{a},\overrightarrow{b}and\overrightarrow{c}$ समतलीय हैं यानी $[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}]=0$

⇒ $(2\overrightarrow{a}\times 3\overrightarrow{b})\cdot 4\overrightarrow{c}=[2\overrightarrow{a}3\overrightarrow{b}4\overrightarrow{c}]$ और $(5\overrightarrow{b}\times 3\overrightarrow{c})\cdot 6\overrightarrow{a}=[5\overrightarrow{b}3\overrightarrow{c}6\overrightarrow{a}]$

⇒ $(2\overrightarrow{a}\times 3\overrightarrow{b})\cdot 4\overrightarrow{c}+(5\overrightarrow{b}\times 3\overrightarrow{c})\cdot 6\overrightarrow{a}=[2\overrightarrow{a}3\overrightarrow{b}4\overrightarrow{c}]+[5\overrightarrow{b}3\overrightarrow{c}6\overrightarrow{a}]$

जैसा कि हम जानते हैं कि, [a b c] = [b c a] = [c a b]

⇒ $(2\overrightarrow{a}\times 3\overrightarrow{b})\cdot 4\overrightarrow{c}+(5\overrightarrow{b}\times 3\overrightarrow{c})\cdot 6\overrightarrow{a}=[2\overrightarrow{a}3\overrightarrow{b}4\overrightarrow{c}]+[6\overrightarrow{a}5\overrightarrow{b}3\overrightarrow{c}]$

जैसा कि हम जानते हैं कि, [λa, b c] = λ [a b c]

⇒ $[2\overrightarrow{a}3\overrightarrow{b}4\overrightarrow{c}]+[6\overrightarrow{a}5\overrightarrow{b}3\overrightarrow{c}]=24[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}]+90[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}]$

जैसा कि हम जानते हैं कि, सदिश $\overrightarrow{a},\overrightarrow{b}और\overrightarrow{c}$ समतलीय केवल तब होता है यदि [abc] = 0

⇒ $[2\overrightarrow{a}3\overrightarrow{b}4\overrightarrow{c}]+[6\overrightarrow{a}5\overrightarrow{b}3\overrightarrow{c}]=0$

⇒ $(2\overrightarrow{a}\times 3\overrightarrow{b})\cdot 4\overrightarrow{c}+(5\overrightarrow{b}\times 3\overrightarrow{c})\cdot 6\overrightarrow{a}=0$

अत: सही विकल्प 3 है।

अवधारणा:

• तीन सदिशों का अदिश त्रिक गुणनफल शून्य होता है यदि उनमें से कोई दो सदिश बराबर हों
• यदि $\overrightarrow{a}$ सदिश $\overrightarrow{b}$ के लंबवत है तो $\overrightarrow{a}\cdot \overrightarrow{b}=0$

गणना :

दिया गया है कि: $\overrightarrow{c}=\overrightarrow{a}+\overrightarrow{b}$ , जहाँ $|\overrightarrow{a}|=|\overrightarrow{b}|\ne 0$

कथन 1: $\overrightarrow{c}$, $(\overrightarrow{a}-\overrightarrow{b})$ के लंबवत है

सबसे पहले $\overrightarrow{c}\cdot (\overrightarrow{a}-\overrightarrow{b})=(\overrightarrow{a}+\overrightarrow{b})\cdot (\overrightarrow{a}-\overrightarrow{b})$ पता करें

⇒ $(\overrightarrow{a}+\overrightarrow{b})\cdot (\overrightarrow{a}-\overrightarrow{b})=|\overrightarrow{a}{|}^2-\overrightarrow{a}\cdot \overrightarrow{b}+\overrightarrow{b}\cdot \overrightarrow{a}-|\overrightarrow{b}{|}^2$

जैसा कि हम जानते हैं कि, $\overrightarrow{a}\cdot \overrightarrow{b}=\overrightarrow{b}\cdot \overrightarrow{a}$

$(\overrightarrow{a}+\overrightarrow{b})\cdot (\overrightarrow{a}-\overrightarrow{b})=|\overrightarrow{a}{|}^2-|\overrightarrow{b}{|}^2$

∵ यह दिया गया है कि $|\overrightarrow{a}|=|\overrightarrow{b}|\ne 0$

⇒ $(\overrightarrow{a}+\overrightarrow{b})\cdot (\overrightarrow{a}-\overrightarrow{b})=|\overrightarrow{a}{|}^2-|\overrightarrow{b}{|}^2=0$

$\overrightarrow{c}\cdot (\overrightarrow{a}-\overrightarrow{b})=0$

अत: कथन 1 सत्य है।

कथन 2: $\overrightarrow{c}$, $\overrightarrow{a}\times \overrightarrow{b}$ लंबवत है

सबसे पहले $\overrightarrow{c}\cdot (\overrightarrow{a}\times \overrightarrow{b})=(\overrightarrow{a}+\overrightarrow{b})\cdot (\overrightarrow{a}\times \overrightarrow{b})$ पता करें

⇒ $(\overrightarrow{a}+\overrightarrow{b})\cdot (\overrightarrow{a}\times \overrightarrow{b})=\overrightarrow{a}\cdot (\overrightarrow{a}\times \overrightarrow{b})+\overrightarrow{b}\cdot (\overrightarrow{a}\times \overrightarrow{b})$

जैसा कि हम जानते हैं कि तीन सदिशों का अदिश त्रिक गुणनफल शून्य होता है, यदि उनमें से कोई भी दो समान हों

$(\overrightarrow{a}+\overrightarrow{b})\cdot (\overrightarrow{a}\times \overrightarrow{b})=0$

अत: कथन 2 सत्य है।

अत: सही विकल्प 3 है।

अवधारणा:

अदिश त्रिक गुणनफल:

यदि $\overrightarrow{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$, 

$\overrightarrow{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$ and,

$\overrightarrow{c}=c_1\hat{i}+c_2\hat{j}+c_3\hat{k}$

तो उनके अदिश त्रिक गुणनफल को

$\overrightarrow{a}\cdot \left(\overrightarrow{b}\times \overrightarrow{c}\right)=\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right|=\left[abc\right]$ के रूप में परिभाषित किया जाता है

गणना:

दिया गया है:

हमारे पास है,

$\overrightarrow{\mathrm{v}}=2\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}$

$\overrightarrow{\mathrm{w}}=\hat{\mathrm{i}}+3\hat{\mathrm{k}}$

$\overrightarrow{\mathrm{v}}\times \overrightarrow{\mathrm{w}}=\left[\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& -1\\ 1& 0& 3\end{array}\right]$

$\Rightarrow \overrightarrow{v}\times \overrightarrow{w}=\hat{i}(3)-\hat{j}(7)+\hat{k}(-1)$

$\Rightarrow \overrightarrow{\mathrm{v}}\times \overrightarrow{\mathrm{w}}=3\hat{\mathrm{i}}-7\hat{\mathrm{j}}-\hat{\mathrm{k}}$

अब, $\overrightarrow{\mathrm{u}}.(\overrightarrow{\mathrm{v}}\times \overrightarrow{\mathrm{w}})=|\overrightarrow{\mathrm{u}}||(3\hat{\mathrm{i}}-7\hat{\mathrm{j}}-\hat{\mathrm{k}})|\cos \theta$

यदि $\overrightarrow{\mathrm{u}}$ एक इकाई सदिश है

$\overrightarrow{\mathrm{u}}.(\overrightarrow{\mathrm{v}}\times \overrightarrow{\mathrm{w}})=1|(3\hat{\mathrm{i}}-7\hat{\mathrm{j}}-\hat{\mathrm{k}})|\cos \theta$

जब, θ = 0°

अधिकतम मान = $\sqrt{3^2+(-7{)}^2+(-1{)}^2}=\sqrt{59}$

संकल्पना:

यदि $\overrightarrow{\mathrm{P}},\overrightarrow{\mathrm{Q}},\overrightarrow{\mathrm{R}}$ सदिश हैं, तो 

• $\overrightarrow{\mathrm{P}}\times \overrightarrow{\mathrm{P}}=0$
• $\overrightarrow{\mathrm{P}}\cdot (\overrightarrow{\mathrm{P}}\times \text{any vector})=0$
• $\overrightarrow{\mathrm{P}}\cdot (\overrightarrow{\mathrm{Q}}\times \overrightarrow{\mathrm{R}})=\overrightarrow{\mathrm{Q}}\cdot (\overrightarrow{\mathrm{R}}\times \overrightarrow{\mathrm{P}})=\overrightarrow{\mathrm{R}}\cdot (\overrightarrow{\mathrm{P}}\times \overrightarrow{\mathrm{Q}})=[\overrightarrow{\mathrm{P}},\overrightarrow{\mathrm{Q}},\overrightarrow{\mathrm{R}}]$
• बिंदु गुणनफल के लिए $(\overrightarrow{\mathrm{P}}+\overrightarrow{\mathrm{Q}})\cdot \overrightarrow{\mathrm{R}}=(\overrightarrow{\mathrm{P}}\cdot \overrightarrow{\mathrm{R}})+(\overrightarrow{\mathrm{Q}}\cdot \overrightarrow{\mathrm{R}})$
• पार गुणनफल के लिए $(\overrightarrow{\mathrm{P}}+\overrightarrow{\mathrm{Q}})\times \overrightarrow{\mathrm{R}}=(\overrightarrow{\mathrm{P}}\times \overrightarrow{\mathrm{R}})+(\overrightarrow{\mathrm{Q}}\times \overrightarrow{\mathrm{R}})$

गणना:

 $\mathrm{S}=(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})\cdot [(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})\times (\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}})]$

⇒ $\mathrm{S}=(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})\cdot [(\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}})]$

⇒ $\mathrm{S}=(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})\cdot [(0+\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}})]$

⇒ $\mathrm{S}=\overrightarrow{\mathrm{a}}\cdot [\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}}]+\overrightarrow{\mathrm{b}}\cdot [\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}}]+\overrightarrow{\mathrm{c}}\cdot [\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}}]$

⇒ $\mathrm{S}=\overrightarrow{\mathrm{a}}\cdot (\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{c}})+\overrightarrow{\mathrm{a}}\cdot (\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{a}})+\overrightarrow{\mathrm{a}}\cdot (\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}})+\overrightarrow{\mathrm{b}}\cdot (\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{c}})+\overrightarrow{\mathrm{b}}\cdot (\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{a}})+\overrightarrow{\mathrm{b}}\cdot (\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}})+\overrightarrow{\mathrm{c}}\cdot (\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{c}})+\overrightarrow{\mathrm{c}}\cdot (\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{a}})+\overrightarrow{\mathrm{c}}\cdot (\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}})$

⇒ $\mathrm{S}=0+0+\overrightarrow{\mathrm{a}}\cdot (\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}})+\overrightarrow{\mathrm{b}}\cdot (\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{c}})+0+0+0+\overrightarrow{\mathrm{c}}\cdot (\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{a}})+0$

⇒ $\mathrm{S}=\overrightarrow{\mathrm{a}}\cdot (\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}})+\overrightarrow{\mathrm{b}}\cdot (\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{c}})+\overrightarrow{\mathrm{c}}\cdot (\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{a}})$

⇒ $\mathrm{S}=\overrightarrow{\mathrm{a}}\cdot (\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}})+\text{boldsymbol}[-\overrightarrow{\mathrm{b}}\cdot (\overrightarrow{\mathrm{c}}\times \overrightarrow{\mathrm{a}})]+[-\overrightarrow{\mathrm{c}}\cdot (\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}})]$

⇒ $\mathrm{S}=[\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}}]-[\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}}]-[\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}}]$

∴ $\text{boldsymbol}\mathrm{S}=-[\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}}]$

प्रयोग की गई संकल्पना :

a, b, और c सदिश  समतलीय हैं जब  $[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}]=0$

$\left|\begin{array}{ccc}\lambda & b_1& c_1\\ a_2& \lambda & c_2\\ a_3& b_3& \lambda \end{array}\right|=0$

गणना:

$\left|\begin{array}{ccc}\lambda & 1& 2\\ 1& \lambda & -1\\ 2& -1& \lambda \end{array}\right|=0$

$\Rightarrow \lambda ({\lambda }^2-1)-1(\lambda +2)+2(-1-2\lambda )=0$

$\Rightarrow {\lambda }^3-\lambda -\lambda -2-2-4\lambda =0$

$\Rightarrow {\lambda }^3-6\lambda -4=0$

$\Rightarrow (\lambda +2)({\lambda }^2-2\lambda -2)=0$

$\Rightarrow \lambda =-2or\lambda =\frac{2\pm \sqrt{12}}{2}$

∴ λ = - 2