Let the image of the point (1,0,7) in the lin | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the line be $L_1: \frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}=\lambda$. Any point $M$ on the line is $(\lambda, 1+2\lambda, 2+3\lambda)$.Let $P =

Vedclass · Accepted Answer

$(3,4,3-2 \sqrt{2})$

Vedclass · Answer

(C) Let the line be $L_1: \frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}=\lambda$. Any point $M$ on the line is $(\lambda, 1+2\lambda, 2+3\lambda)$.Let $P = (1,0,7)$. The vector $\overrightarrow{PM} = (\lambda-1, 1+2\lambda, 3\lambda-5)$.Since $\overrightarrow{PM}$ is perpendicular to the line $L_1$ with direction vector $\vec{b} = (1,2,3)$,we have $\overrightarrow{PM} \cdot \vec{b} = 0$.$(\lambda-1)(1) + (1+2\lambda)(2) + (3\lambda-5)(3) = 0 \Rightarrow \lambda-1+2+4\lambda+9\lambda-15 = 0 \Rightarrow 14\lambda = 14 \Rightarrow \lambda = 1$.Thus,$M = (1, 3, 5)$.Since $M$ is the midpoint of $PQ$,where $Q = (\alpha, \beta, \gamma)$,we have $M = \frac{P+Q}{2} \Rightarrow Q = 2M - P = 2(1,3,5) - (1,0,7) = (2-1, 6-0, 10-7) = (1,6,3)$.So,$(\alpha, \beta, \gamma) = (1,6,3)$.Let the direction cosines of the required line be $(l, m, n)$. Given $m = \cos(\frac{2\pi}{3}) = -\frac{1}{2}$ and $n = \cos(\frac{3\pi}{4}) = -\frac{1}{\sqrt{2}}$.Since $l^2+m^2+n^2=1$,$l^2 + (-\frac{1}{2})^2 + (-\frac{1}{\sqrt{2}})^2 = 1 \Rightarrow l^2 + \frac{1}{4} + \frac{1}{2} = 1 \Rightarrow l^2 = \frac{1}{4}$.Since the line makes an acute angle with the $x$-axis,$l = \frac{1}{2}$.The line passes through $(1,6,3)$ with direction vector $(\frac{1}{2}, -\frac{1}{2}, -\frac{1}{\sqrt{2}})$,or $(1, -1, -\sqrt{2})$.The equation is $\frac{x-1}{1} = \frac{y-6}{-1} = \frac{z-3}{-\sqrt{2}} = \mu$.For $\mu = 2$,$x = 3, y = 4, z = 3-2\sqrt{2}$. This matches option $C$.

Similar Questions

The angle between the lines $\frac{x-1}{l}=\frac{y+1}{m}=\frac{z}{n}$ and $\frac{x+1}{m}=\frac{y-3}{n}=\frac{z-1}{l}$,where $l > m > n$ and $l, m, n$ are roots of the equation $x^3+x^2-4x-4=0$,is

The shortest distance between the lines $\frac{x - 3}{3} = \frac{y - 8}{-1} = \frac{z - 3}{1}$ and $\frac{x + 3}{-3} = \frac{y + 7}{2} = \frac{z - 6}{4}$ is

If $A(1, 0, 2)$,$B(2, 1, 0)$,$C(2, -5, 3)$,and $D(0, 3, 2)$ are four points and the point of intersection of the lines $AB$ and $CD$ is $P(a, b, c)$,then $a + b + c =$

The shortest distance between the lines $r=(-2 \hat{i}+\hat{j}-\hat{k})+r(2 \hat{i}+3 \hat{j}-\hat{k})$ and $r=(\hat{i}-\hat{j}+2 \hat{k})+k(-\hat{i}+2 \hat{j}+4 \hat{k})$ is

The distance of the point having position vector $-\hat{i} + 2\hat{j} + 6\hat{k}$ from the straight line passing through the point $(2, 3, -4)$ and parallel to the vector $6\hat{i} + 3\hat{j} - 4\hat{k}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module