Perpendicular Lines MCQ Quiz in मल्याळम - Objective Question with Answer for Perpendicular Lines - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept Used:

If two lines with direction ratios (a₁, b₁, c₁) and (a₂, b₂, c₂) are perpendicular, then a₁a₂ + b₁b₂ + c₁c₂ = 0.

Calculation:

Given:

Line 1: \(\frac{1-x}{3} = \frac{7y-14}{2p} = \frac{z-3}{2}\)

Line 2: \(\frac{7-7x}{3p} = \frac{y-5}{1} = \frac{6-z}{5}\)

Lines are at right angles.

Line 1: \(\frac{x-1}{-3} = \frac{y-2}{\frac{2p}{7}} = \frac{z-3}{2}\)

Line 2: \(\frac{x-1}{-\frac{3p}{7}} = \frac{y-5}{1} = \frac{z-6}{-5}\)

Direction ratios of Line 1: (-3, \(\frac{2p}{7}\), 2)

Direction ratios of Line 2: (\(\frac{-3p}{7}\), 1, -5)

Since the lines are perpendicular:

⇒ \((-3)\left(-\frac{3p}{7}\right) + \left(\frac{2p}{7}\right)(1) + (2)(-5) = 0\)

⇒ \(\frac{9p}{7} + \frac{2p}{7} - 10 = 0\)

⇒ \(\frac{11p}{7} = 10\)

 ⇒\(p = \frac{70}{11}\)

Hence option 3 is correct

Calculation

Direction ratios of line RQ:

(2-0, -3-(-1), -1-3) = (2, -2, -4) = (1, -1, -2)

Equation of line RQ in parametric form:

\(x = 0 + \lambda(1) = \lambda\)

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

\(z = 3 + \lambda(-2) = 3 - 2\lambda\)

Let the foot of the perpendicular from P on line RQ be F(\(\lambda\), -1-\(\lambda\), 3-2\(\lambda\)).

Direction ratios of PF:

(\(\lambda\)-1, -1-\(\lambda\)-8, 3-2\(\lambda\)-4) = (\(\lambda\)-1, -9-\(\lambda\), -1-2\(\lambda\))

Since PF is perpendicular to RQ, the dot product of their direction ratios is zero:

⇒ 1(\(\lambda\)-1) - 1(-9-\(\lambda\)) - 2(-1-2\(\lambda\)) = 0

⇒ \(\lambda\) - 1 + 9 + \(\lambda\) + 2 + 4\(\lambda\) = 0

⇒ 6\(\lambda\) + 10 = 0

⇒ 6\(\lambda\) = -10

⇒ \(\lambda\) = -10/6 = -5/3

Coordinates of F:

\(x = \lambda = -5/3\)

\(y = -1 - \lambda = -1 - (-5/3) = -1 + 5/3 = 2/3\)

\(z = 3 - 2\lambda = 3 - 2(-5/3) = 3 + 10/3 = 19/3\)

∴ The coordinates of the foot of the perpendicular are (-5/3, 2/3, 19/3).

Hence option 3 is correct

Concept:

Then invariant point of a transformation w = T(z) is given by z = T(z).

Explanation:

The invariant points of \(\rm w=\frac{2Z-4}{Z+2}\) is given by

\(\rm Z=\frac{2Z-4}{Z+2}\)

⇒ Z² + 2Z = 2Z - 4

⇒ Z² = -4

⇒ Z = ± 2i

(3) is true.

Relation :\(\dfrac{x-x_{1}}{a}=\dfrac{y-y_{1}}{b}=\dfrac{-\left ( ax_{1}+by_{1}+cz_{1} \right )}{a^{2}+b^{2}}\)

Here \((x_1,y_1)\)=(0,0)

\(\dfrac{x-0}{3}=\dfrac{y-0}{1}=\dfrac{-\left ( \left ( 3\times 0 \right )+\left ( 1\times 0 \right )-\lambda \right )}{3^{2}+1^{2}}\)

\(x=\dfrac{3\lambda }{10}\) and \(y=\dfrac{\lambda }{10}\)

Hence foot of perpendicular \(P=\left ( \dfrac{3\lambda }{10},\dfrac{\lambda }{10} \right )\)

Line meets X-axis at \(A=\left ( \dfrac{\lambda }{3},0 \right )\)

\(BP=\sqrt{\left ( \dfrac{3\lambda }{10} \right )^{2}+\left ( \dfrac{\lambda }{10}-\lambda \right )^{2}}\)

\(\Rightarrow BP=\sqrt{\dfrac{9\lambda ^{2}}{100}+\dfrac{81\lambda ^{2}}{100}}\)

\(\therefore BP=\sqrt{\dfrac{90\lambda ^{2}}{100}}\)

\(AP=\sqrt{\left ( \dfrac{\lambda }{3}-\dfrac{3\lambda }{10} \right )^{2}+\left ( 0-\dfrac{\lambda }{10} \right )^{2}}\)

\(\Rightarrow AP=\sqrt{\dfrac{\lambda ^{2}}{900}+\dfrac{\lambda ^{2}}{100}}\)

\(\therefore AP=\sqrt{\dfrac{10\lambda ^{2}}{900}}\)

\(\therefore BP:AP=9:1\)

Hence, correct option is 'A'.

Equation of \(L_1\) is
\( y = - \dfrac{1}{2} x + \dfrac{5}{2}\) ....(1)
Equation of \(L_2\) is
\(y = 2x - 5\) ...(2)
By (1) and (2)
\(x = 3\)
\(y = 1 \Rightarrow h = 3, k= 1\)
\(\dfrac{k}{h} = \dfrac{1}{3}\)

\( \left(2,\dfrac{-3}{2},\dfrac{1}{2}\right) \)

Now, \( \overset{\rightarrow }{MP}.(10\hat{i}-7\hat{j}+\hat{k})=0 \)

\( \Rightarrow \lambda=\dfrac{1}{2} \)

\( \therefore \) Length of perpendicular

\( (=PM)=\sqrt{0+\dfrac{1}{4}+\dfrac{49}{4}} \)

\( =\sqrt{\dfrac{50}{4}}=\sqrt{\dfrac{25}{2}}=\sqrt{\dfrac{5}{\sqrt{2}}} \), which is greater than \( 3 \) but \( less than 4 \)

Slope of given line is \(\dfrac{2}{3}\)

Lines are perpendicular so

\(\dfrac{17 - \beta}{-8} \times \dfrac{2}{3} = -1\)

\(\beta = 5\)

Calculation

Given 

x = y, z = 1 ⇒ L₁ : \(\frac{x}{1} = \frac{y}{1} = \frac{z - 1}{0} = λ\)

⇒ Q = (λ, λ, 1)

⇒ \(\vec {PQ} = (λ-a) \hat i +(λ-a) \hat j + (1-a)\hat k\)

\(\vec {PQ} \perp L_1\) ⇒  \(((λ-a) \hat i +(λ-a) \hat j + (1-a)\hat k)\cdot(\hat i+\hat j +0\hat k)\) = 0

x = –y, z = –1 ⇒ L₂ : \(\frac{x}{1} =\frac{y}{-1}=\frac{z+1}{0}=k\)

⇒ R = (k, -k, -1)

⇒ \(\vec {PR} = (k-a) \hat i +(-k-a) \hat j + (-1-a)\hat k\)

\(\vec {PR} \perp L_2\) ⇒ \(((k-a) \hat i +(-k-a) \hat j + (-1-a)\hat k)\cdot(\hat i-\hat j +0\hat k)\) = 0

⇒ k - a + k + a  = 0 ⇒ k = 0

∠QPR is a right angle ⇒ \(\vec {PQ} \perp \vec {PR}\)

⇒ \( ((λ-a) \hat i +(λ-a) \hat j + (1-a)\hat k)\cdot((k-a) \hat i +(-k-a) \hat j + (-1-a)\hat k) = 0\)

⇒ \( ((λ-a)(k-a)+(λ-a)(-k-a)+ (1-a)(-1-a) = 0\)

⇒ (1-a)(1+a) = 0 ⇒ a² = 1

⇒12a2 = 12

Concept:

Let two lines having direction ratio’s a_1, b₁, c₁ and a_2, b₂, c₂ respectively.

Condition for perpendicular lines: a₁a₂ + b₁b₂ + c₁c₂ = 0

Calculation:

Direction ratio’s of two lines are given as 1, 2, p − 1 and -3, 1, 2

Lines are perpendicular,

∴ 1 × -3 + 2 × 1 + (p – 1) × 2 = 0

⇒ -3 + 2 + 2p – 2 = 0

⇒ 2p = 3

∴ p = 3/2

Concept:

Let the two lines have direction ratio’s a1, b1, c1 and a2, b2, c2 respectively.

Condition for perpendicular lines: a1a2 + b1b2 + c1c2 = 0

Calculation:

Given lines are  \(\frac{{{\rm{x}} + 4}}{{2{\rm{}}}} = \frac{{4 - {\rm{y}}}}{-2} = \frac{{{\rm{2z}} - 4}}{{\rm 2k}}\) and \(\frac{{{\rm{x}} +3}}{\rm -k} = \frac{{\rm{y-3}}}{{\rm{2}}} = \frac{{{\rm{z}} + 1}}{5}\) 

Write the above equation of a line in the standard form of lines

\( \Rightarrow \frac{{{\rm{x}} + 4}}{{2{\rm{}}}} = \frac{-{(\rm y - {\rm{4})}}}{-2} = \frac{2{{(\rm{z}} - 2)}}{{\rm 2k}} \Leftrightarrow \frac{{\left( {{\rm{x}} +4 } \right)}}{{\rm{2}}} = \frac{{{\rm{y}} - 4}}{{ 2}} = \frac{{{\rm{z}} - 2}}{{ \rm k}}\)

So, the direction ratio of the first line is (2, 2, k)

\(\frac{{{\rm{x}} +3}}{\rm -k} = \frac{{\rm{y-3}}}{{\rm{2}}} = \frac{{{\rm{z}} + 1}}{5}\)

So, direction ratio of second line is (-k, 2, 5)

Lines are perpendicular,

∴ (2 × -k) + (2 × 2) + (k × 5) = 0

⇒ -2k + 4 + 5k = 0

⇒ 3k + 4 = 0

∴ k = -4/3