The angle between the line vec{r} = (hat{i} + | vedclass

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

Question

(A) The direction vector of the line is $\vec{b} = 2\hat{i} - \hat{j} + \hat{k}$.The normal vector to the plane is $\vec{n} = \hat{i} + \hat{j} + 3\ha

Vedclass · Accepted Answer

$\cos^{-1} (\frac{4}{\sqrt{66}})$

Vedclass · Answer

(A) The direction vector of the line is $\vec{b} = 2\hat{i} - \hat{j} + \hat{k}$.The normal vector to the plane is $\vec{n} = \hat{i} + \hat{j} + 3\hat{k}$.The angle $	heta$ between the line and the normal to the plane is given by the formula $\cos 	heta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$.Calculating the dot product: $\vec{b} \cdot \vec{n} = (2)(1) + (-1)(1) + (1)(3) = 2 - 1 + 3 = 4$.Calculating the magnitudes: $|\vec{b}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$ and $|\vec{n}| = \sqrt{1^2 + 1^2 + 3^2} = \sqrt{1 + 1 + 9} = \sqrt{11}$.Thus,$\cos 	heta = \frac{4}{\sqrt{6} \cdot \sqrt{11}} = \frac{4}{\sqrt{66}}$.Therefore,$	heta = \cos^{-1} (\frac{4}{\sqrt{66}})$.

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

Similar Questions

Let $\Pi$ be a plane containing the points $(0,-5,-1), (1,-2,5), (-3,5,0)$ and $L$ be a line passing through the point $(0,-5,-1)$ and parallel to the vector $\hat{i}+5\hat{j}-6\hat{k}$. Then the length of the projection of the unit normal vector to the plane $\Pi$ on the line $L$ is

If $P=(0,1,0)$ and $Q=(0,0,1)$,then the length of the projection of the line segment $PQ$ on the plane $x+y+z=3$ is:

Let $P$ be a plane containing the line $\frac{x-1}{3}=\frac{y+6}{4}=\frac{z+5}{2}$ and parallel to the line $\frac{x-3}{4}=\frac{y-2}{-3}=\frac{z+5}{7}$. If the point $(1, -1, \alpha)$ lies on the plane $P$,then the value of $|5\alpha|$ is equal to ....... .

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:

Find the ratio in which the plane $2x + 3y + 5z = 1$ divides the line segment joining the points $(1, 0, -3)$ and $(1, -5, 7)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module