Scalar and Vector Product MCQ Quiz in मल्याळम - Objective Question with Answer for Scalar and Vector Product - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

Scalar Product of Two Vectors - \(\rm \overrightarrow{a} .\overrightarrow{b} = |\overrightarrow{a}| |\overrightarrow{b}| cos\ θ \)

Vector Product of Two Vectors - \(\rm \overrightarrow{a} \times \overrightarrow{b} = |\overrightarrow{a}| |\overrightarrow{b}| sin\ θ \: \widehat{n} \)

\(\: \widehat{n}\) is the unit vector perpendicular vector, θ being the angle between  \(\rm \overrightarrow{a} \) and \(\rm \overrightarrow{b} \)

Calculation:

Given

\(\rm |\overrightarrow{a} .\overrightarrow{b}| = |\overrightarrow{a} \times \overrightarrow{b}|\)

⇒ \(\rm |\overrightarrow{a}| |\overrightarrow{b}| \cos θ = |\overrightarrow{a}| |\overrightarrow{b}| \sin θ \)

⇒ cos θ = sin θ

⇒ tan θ = 1

⇒ θ = \(\rm \frac{\pi}{4}\)

The angle between  \(\rm \overrightarrow{a} \) and \(\rm \overrightarrow{b} \) is \(\rm \frac{\pi}{4}\)

Calculation:

Given: In the right-angled triangle ABC, hypotenuse AC = p

As we can see from the above diagram angle between vector \(\overrightarrow {BC} \;and\;\overrightarrow {BA}\) is 90° ⇒ \(\overrightarrow {BC} \; \cdot \overrightarrow {BA} = 0\)

\(\Rightarrow \;\overrightarrow {AB} \cdot \overrightarrow {AC} + \overrightarrow {BC} \; \cdot \overrightarrow {BA} + \overrightarrow {CA} \cdot \overrightarrow {CB} = \;\overrightarrow {AB} \cdot \overrightarrow {AC} + \;\overrightarrow {CA} \cdot \overrightarrow {CB}\)

\(\Rightarrow \;\overrightarrow {AB} \cdot \overrightarrow {AC} + \overrightarrow {BC} \; \cdot \overrightarrow {BA} + \overrightarrow {CA} \cdot \overrightarrow {CB} = \overrightarrow {AB} \cdot \overrightarrow {AC} + \;\overrightarrow {AC} \cdot \overrightarrow {BC}\)

\( = \;\overrightarrow {AC} \cdot \left( {\overrightarrow {AB} + \;\overrightarrow {BC} } \right) = \overrightarrow {AC} \cdot \;\overrightarrow {AC} = {\left| {\overrightarrow {AC} } \right|^2}\)

Concept:

Vector Triple Product: Vector Triple Product is a vector quantity.

Vector triple product of three vectors a, b, c is defined as the cross product of vector a with the cross product of vectors b and c, i.e. a × (b × c)

a × (b × c) = (a . c) b – (a . b) c

a.b = |a||b|cos θ

Calculation: 

Here, |a| = 1, |b| = 1, |c| = 2

\(\rm \vec a × (\vec a × \vec c) - \vec b = 0\)

\(\rm (\vec a.\vec c)\vec a-(\vec a.\vec a)\vec c-\vec b=0\)

\( \rm (\vec a.\vec c)\vec a=(\vec a.\vec a)\vec c+\vec b\)

\(\rm (|a||c|cosθ)\vec a=(|a||a|cos 0)\vec c+\vec b\)

\(\rm 2\cos θ\; \vec{a} = \vec{c}+\vec{b}\)

\(\rm 2\cos θ\; \vec{a} - \vec{c}=\vec{b}\)

Taking magnitude both sides, we get

4cos² θ |a|² + |c|² - 2 × 2cos θ \(\rm \vec a \cdot \vec c\) = |b|²

4cos2 θ + 4 - 2 × 2cos θ × |a||c|cos θ = 1

4cos2 θ + 4 - 8cos2 θ  = 1

4cos2 θ = 3

cos2 θ = \(\frac 3 4\)

cos θ = \(\frac {\sqrt{3}}{2}\)

∴ θ = π / 6

Concept:

For two vectors \(\rm \rm \vec A\) and \(\rm \vec B\) at an angle θ to each other:

• Dot Product is defined as \(\rm \vec A.\vec B=\left|\vec A\right|\left|\vec B\right|\cos \theta\).
• Cross Product is defined as \(\rm \vec A\times \vec B=\hat n\left|\vec A\right|\left|\vec B\right|\sin \theta\) where \(\rm \hat n\) is the unit vector perpendicular to the plane containing \(\rm \rm \vec A\) and \(\rm \vec B\).

Calculation:

According to the question:

\(\rm \vec a.\vec b=\left|\vec a\right|\left|\vec b\right|\cos \theta=3\)                 ...(1)

\(\rm |\vec a\times\vec b|=|\hat n\left|\vec a\right|\left|\vec b\right|\sin\theta|=3\sqrt3\)       

As we know, \(\rm | \hat n| = 1\)         

\(\rm \left|\vec a\right|\left|\vec b\right|\sin\theta=3\sqrt3\)                      ...(2)

Dividing (2) by (1), we get:

tan θ = √3

⇒ θ = \(\rm \frac{\pi}{3}\).

Concept:

\(\rm \vec a.\vec b=|\vec a||\vec b|\cos\theta \)

Calculation:

Here, \(\vec a + \vec b + \vec c = 0\)

\(\Rightarrow \vec b + \vec c = -\vec a\)

Taking magnitude and squaring both sides,

\(\rm \Rightarrow | \vec b + \vec c|^2 = |-\vec a|^2\)

\(\rm \Rightarrow |\vec b|^2+|\vec c|^2+2\vec b.\vec c =100\)

\(\rm \Rightarrow 2 |\vec b|.|\vec c|cos θ =100-(16 + 36)\)

\(\\ \rm \Rightarrow cosθ =\frac{48}{2\;\times\; 4 \;\times\; 6}\)

\(\Rightarrow θ =cos^{-1}(1)\)

θ = 0°

Hence, option (1) is correct. 

Concept:

a.b = |a||b|cosθ

Calculation: 

Here,  \(\rm \vec a = \left(\vec i + 2\vec j - 3\vec k\right)\) and \(\vec b = \left(3\vec i - \vec j + 2\vec k\right)\) 

\(\rm \left(\vec a + \vec b\right) = (\vec i + 2\vec j - 3\vec k )+(\vec 3i -\vec j +2\vec k)\\=\vec 4i + \vec j - \vec k\\ |\rm (\vec a + \vec b)|=\sqrt{4^2+1^2+(-1)^2}\\ =\sqrt{18}\\=3\sqrt2\)

\(\rm \left(\vec a - \vec b\right) = (\vec i + 2\vec j - 3\vec k )-(\vec 3i -\vec j +2\vec k)\\ =-\vec 2i + \vec3 j - \vec 5 k\\ |\rm (\vec a - \vec b)|=\sqrt{(-2)^2+3^2+(-5)^2}\\ =\sqrt{38}\\\)

Now, 

\(\rm \left(\vec a +\vec b\right) .\left(\vec a - \vec b\right) =(\vec 4i + \vec j - \vec k) \cdot (-\vec 2i + \vec3 j - \vec 5 k)= -8+3+5\\ |\left(\vec a + \vec b\right)||\left(\vec a - \vec b\right)|cosθ =0\\ \Rightarrow θ = \frac \pi 2\)

Concept:

• If \(\vec a\;and\;\vec b\) are two vectors, then the scalar product between the given vectors is given by: \(\vec a \cdot \;\vec b = \left| {\vec a} \right| \times \left| {\vec b} \right| \times \cos θ \)
• If \(\vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k\;and\;\vec b = {b_1}\hat i + {b_2}\hat j + {b_3}\hat k\) then \(\vec a \cdot \;\vec b = {a_1}{b_1} + {a_2}{b_2} + {a_3}{b_3}\)
• If \(\vec a = \;{a_1}\hat i + {a_2}\hat j + {a_3}\hat k\) is a vector then the magnitude of the vector is given by \(\left| {\vec a} \right| = \sqrt {a_1^2 + a_2^2 + a_3^2} \;\)

Given: \(\vec a = 3\hat i - 2\hat j + \;\hat k\;and\;\vec b = \;\hat i - 2\hat j - 3\hat k\)

As we know that, \(\vec a \cdot \;\vec b = {a_1}{b_1} + {a_2}{b_2} + {a_3}{b_3}\)

\(\vec a \cdot \;\vec b = 3 + 4 - 3 = 4\)

As we know that, if \(\vec a = \;{a_1}\hat i + {a_2}\hat j + {a_3}\hat k\) is a vector then the magnitude of the vector is given by \(\left| {\vec a} \right| = \sqrt {a_1^2 + a_2^2 + a_3^2} \;\)

Concept:

Let \(\vec{u}\) be a vector then \({{\left| {\vec{u}} \right|}^{2}}=~\vec{u}\cdot ~\vec{u}\)

Calculation:

\(\left( \vec{a}+\vec{b} \right)\cdot \left( \vec{a}-\vec{b} \right)=\\{{\left| {\vec{a}} \right|}^{2}}-\left| {\vec{a}} \right|\times \left| {\vec{b}} \right|\times \cos \theta +\left| {\vec{a}} \right|\times \left| {\vec{b}} \right|\times \cos \theta -{{\left| {\vec{b}} \right|}^{2}}=~{{\left| {\vec{a}} \right|}^{2}}-{{\left| {\vec{b}} \right|}^{2}}\)

Hence, statement 1 is true.

\(\left( \left| \vec{a}+\vec{b} \right| \right)\left( \left| \vec{a}-\vec{b} \right| \right)=\\\sqrt{{{\left| {\vec{a}} \right|}^{2}}+{{\left| {\vec{b}} \right|}^{2}}+2\times \left| {\vec{a}} \right|\times \left| {\vec{b}} \right|\times \cos \theta }~\times ~\sqrt{{{\left| {\vec{a}} \right|}^{2}}+{{\left| {\vec{b}} \right|}^{2}}-2\times \left| {\vec{a}} \right|\times \left| {\vec{b}} \right|~\times \cos \theta }\)

If cos θ = 1

\(\Rightarrow \left( \left| \vec{a}+\vec{b} \right| \right)\left( \left| \vec{a}-\vec{b} \right| \right)={{\left| {\vec{a}} \right|}^{2}}-{{\left| {\vec{b}} \right|}^{2}}\)

But if cos θ ≠ 1

\(\Rightarrow \left( \left| \vec{a}+\vec{b} \right| \right)\left( \left| \vec{a}-\vec{b} \right| \right)\ne {{\left| {\vec{a}} \right|}^{2}}-{{\left| {\vec{b}} \right|}^{2}}\)

Hence, statement 2 is not true .

\({{\left| \vec{a}\cdot \vec{b} \right|}^{2}}+{{\left| \vec{a}\times \vec{b} \right|}^{2}}=\\{{\left| {\vec{a}} \right|}^{2}}\times {{\left| {\vec{b}} \right|}^{2}}\times {{\cos }^{2}}\theta +~{{\left| {\vec{a}} \right|}^{2}}\times {{\left| {\vec{b}} \right|}^{2}}\times {{\sin }^{2}}\theta =~{{\left| {\vec{a}} \right|}^{2}}{{\left| {\vec{b}} \right|}^{2}}\)

Concept:

If vector v = ai + bj + ck, then magnitude of vector v is \( √ {a^2+b^2+c^2}\).

A vector \(\overrightarrow{v} \) in the plane of \(\overrightarrow{a} \) and \(\overrightarrow{b} \) is \(\overrightarrow{v}=\overrightarrow{a}+\lambda \overrightarrow{b}\)

Calculation:

A vector \(\overrightarrow{v} \) in the plane of \(\overrightarrow{a} \) and \(\overrightarrow{b} \) is  

\(\overrightarrow{v}=\overrightarrow{a}+\lambda \overrightarrow{b}\)

⇒ \(\overrightarrow{v} = (\hat i + \hat j + \hat k)+\lambda(\hat i-\hat j + \hat k)\)

⇒ \(\overrightarrow{v} \) = \({(1+\lambda)\hat i+(1-\lambda)\hat j+(1+\lambda)\hat k}\)

Projection of \(\overrightarrow{v} \) on \(\overrightarrow{c} = \frac{1}{√3}\)(given)

⇒ \(\frac{\overrightarrow{v}.\overrightarrow{c}}{\mid\overrightarrow{c}\mid} = \frac{1}{√3}\)

⇒ \(\frac{(1+\lambda)-(1-\lambda)-(1+\lambda)}{√3}=\frac{1}{√3}\)            [\( ​​​​\mid \overrightarrow{c}\mid =\sqrt{{1^2+1^2+1^2}}\) = √3]

⇒ \((1+\lambda)-(1-\lambda)-(1+\lambda)=1 \)

⇒ \(\lambda=2 \)

 \(\overrightarrow{v}=3\hat i -\hat j +3 \hat k\)

∴ Vector \(\overrightarrow{v}=3\hat i -\hat j +3 \hat k\)

Concept:

I. If \(\vec a\;and\;\vec b\) are two vectors, then the scalar product between the given vectors is given by:

\(\vec a \cdot \;\vec b = \left| {\vec a} \right| \times \left| {\vec b} \right| \times \cos \theta \)

II. If \(\vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k\;and\;\vec b = {b_1}\hat i + {b_2}\hat j + {b_3}\hat k\) then \(\vec a \times \;\vec b = \;\;\left| {\vec a} \right| \times \left| {\vec b} \right| \times \sin \theta \times \;\hat n,\;where\;\hat n\) is the unit vector perpendicular to both \(\vec a\;and\;\vec b\)

Calculation:

Given: \(\left| {\vec a} \right| = 2,\;\left| {\vec b} \right| = 5\;and\left| {\vec a \times \;\vec b} \right| = 8\)

As we know that, If \(\vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k\;and\;\vec b = {b_1}\hat i + {b_2}\hat j + {b_3}\hat k\) then \(\vec a \times \;\vec b = \;\;\left| {\vec a} \right| \times \left| {\vec b} \right| \times \sin \theta \times \;\hat n,\;where\;\hat n\) is the unit vector perpendicular to both \(\vec a\;and\;\vec b\)

\(\left| {\vec a \times \;\vec b} \right| = \left| {\vec a} \right| \times \left| {\vec b} \right| \times \left| {\sin \theta } \right| \times \left| {\hat n} \right| = \;\left| {\vec a} \right| \times \left| {\vec b} \right| \times \sin \theta \)

\( \Rightarrow \sin \theta = \frac{{\left| {\vec a \times \;\vec b} \right|}}{{\left| {\vec a} \right| \times \left| {\vec b} \right|}} = \frac{8}{{10}} = \frac{4}{5}\)

\( \Rightarrow \cos \theta = \frac{3}{5}\)

\( \Rightarrow \cos \theta \; = \;\frac{3}{5}\)

As we know that, If \(\vec a\;and\;\vec b\) are two vectors, then the scalar product between the given vectors is given by:  \(\vec a \cdot \;\vec b = \left| {\vec a} \right| \times \left| {\vec b} \right| \times \cos \theta \)

\( \Rightarrow \vec a \cdot \;\vec b = 2 \times 5 \times \frac{3}{5} = 6\)