L is a line passing through the point A(1, 0, | vedclass

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

Question

(B) The equation of line $L$ is $\frac{x-1}{0} = \frac{y-0}{1} = \frac{z+3}{-2} = \lambda$. Any point $P$ on $L$ is $(1, \lambda, -2\lambda - 3)$. The

Vedclass · Accepted Answer

$y - 2z + 4 = 0$

Vedclass · Answer

(B) The equation of line $L$ is $\frac{x-1}{0} = \frac{y-0}{1} = \frac{z+3}{-2} = \lambda$. Any point $P$ on $L$ is $(1, \lambda, -2\lambda - 3)$. The distance $d$ of point $P$ from the plane $2x + 3y + 5z - 1 = 0$ is given by: $d = \frac{|2(1) + 3(\lambda) + 5(-2\lambda - 3) - 1|}{\sqrt{2^2 + 3^2 + 5^2}} = \frac{|2 + 3\lambda - 10\lambda - 15 - 1|}{\sqrt{38}} = \frac{|-7\lambda - 14|}{\sqrt{38}}$. For minimum distance,$-7\lambda - 14 = 0$,which gives $\lambda = -2$. Substituting $\lambda = -2$ into the coordinates of $P$,we get $P(1, -2, 1)$. The vector $\vec{AP} = P - A = (1-1, -2-0, 1-(-3)) = (0, -2, 4)$. The equation of the plane passing through $P(1, -2, 1)$ with normal vector $\vec{n} = (0, -2, 4)$ is: $0(x - 1) - 2(y + 2) + 4(z - 1) = 0$. $-2y - 4 + 4z - 4 = 0 \Rightarrow -2y + 4z - 8 = 0 \Rightarrow y - 2z + 4 = 0$.

$L$ is a line passing through the point $A(1, 0, -3)$ and parallel to a line having direction ratios $0, 1, -2$. $P$ is a point on the line $L$ which is at a minimum distance from the plane $2x + 3y + 5z = 1$. Then,the equation of the plane through $P$ and perpendicular to $AP$ is

Similar Questions

Let $N$ be the foot of the perpendicular from the point $P(1, -2, 3)$ on the line passing through the points $(4, 5, 8)$ and $(1, -7, 5)$. Then the distance of $N$ from the plane $2x - 2y + z + 5 = 0$ is $.......$.

Let the line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lie in the plane $x+3y-\alpha z+\beta=0$. Then the value of $(\beta-\alpha)$ is equal to

The distance between the line $\vec{r} = (2\hat{i} - 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - \hat{j} + 4\hat{k})$ and the plane $\vec{r} \cdot (\hat{i} + 5\hat{j} + \hat{k}) = 5$ is:

The ratio in which the line segment joining the points $(2, -4, 3)$ and $(-4, 5, -6)$ is divided by the plane $3x + 2y + z - 4 = 0$ is:

The angle between the line $\bar{r}=(\hat{i}+2\hat{j}-\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})$ and the plane $\bar{r} \cdot (2\hat{i}-\hat{j}+\hat{k})=4$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module