If d is the distance of the point (2, 5, 10) | vedclass

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

Question

(C) Step $1$: Identify the normal vector $\vec{n}$ to the plane. The lines are parallel as they have the same direction vector $\vec{b} = \hat{i} + 2\

Vedclass · Accepted Answer

$90$

Vedclass · Answer

(C) Step $1$: Identify the normal vector $\vec{n}$ to the plane. The lines are parallel as they have the same direction vector $\vec{b} = \hat{i} + 2\hat{j} - 2\hat{k}$.Step $2$: Let $A = (0, 4, -1)$ and $B = (2, 1, 0)$ be points on the lines. The vector $\vec{AB} = (2-0)\hat{i} + (1-4)\hat{j} + (0-(-1))\hat{k} = 2\hat{i} - 3\hat{j} + \hat{k}$.Step $3$: The normal vector $\vec{n} = \vec{AB} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -3 & 1 \ 1 & 2 & -2 \end{vmatrix} = \hat{i}(6-2) - \hat{j}(-4-1) + \hat{k}(4+3) = 4\hat{i} + 5\hat{j} + 7\hat{k}$.Step $4$: The equation of the plane is $4(x-2) + 5(y-1) + 7(z-0) = 0$, which simplifies to $4x + 5y + 7z - 13 = 0$.Step $5$: The distance $d$ from $(2, 5, 10)$ to the plane is $d = \frac{|4(2) + 5(5) + 7(10) - 13|}{\sqrt{4^2 + 5^2 + 7^2}} = \frac{|8 + 25 + 70 - 13|}{\sqrt{16 + 25 + 49}} = \frac{90}{\sqrt{90}} = \sqrt{90}$.Step $6$: Therefore, $d^2 = 90$.

If $d$ is the distance of the point $(2, 5, 10)$ from the plane containing the lines $\vec{r} = (4\hat{j} - \hat{k}) + \lambda(\hat{i} + 2\hat{j} - 2\hat{k})$ and $\vec{r} = (2\hat{i} + \hat{j}) + \mu(\hat{i} + 2\hat{j} - 2\hat{k})$, then $d^2 = $

Similar Questions

The line of intersection of the planes $r \cdot (i - 3j + k) = 1$ and $r \cdot (2i + 5j - 3k) = 2$ is parallel to which vector?

The angle $\theta$ between the line $x - 1 = \frac{y - 2}{-1} = \frac{z - 3}{2}$ and the plane $\vec{r} \cdot (2\hat{i} + \hat{j} + \hat{k}) = 10$ is

$A$ plane which passes through the point $(3, 2, 0)$ and the line $\frac{x - 3}{1} = \frac{y - 6}{5} = \frac{z - 4}{4}$ is

Let $L$ be the line of intersection of the planes $2x + 3y + z = 1$ and $x + 3y + 2z = 2$. If $L$ makes an angle $\alpha$ with the positive direction of the $x$-axis,then $\cos \alpha$ is:

The line $\frac{x - 1}{2} = -(y + 1) = \frac{z}{3}$ and the plane $3x + 2y - z = 5$ intersect at a point. The coordinates of the point are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module