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

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

Question

(C) The angle $	heta$ between a line with direction vector $\vec{b}$ and a plane with normal vector $\vec{n}$ is given by $\sin 	heta = \left|\frac{

Vedclass · Accepted Answer

$\sin^{-1}\left(\frac{\sqrt{5}}{2\sqrt{7}}\right)$

Vedclass · Answer

(C) The angle $	heta$ between a line with direction vector $\vec{b}$ and a plane with normal vector $\vec{n}$ is given by $\sin 	heta = \left|\frac{\vec{b} \cdot \vec{n}}{|\vec{b}| |\vec{n}|}ight|$.Here,the direction vector of the line is $\vec{b} = 3\hat{i} + \hat{j}$ and the normal vector to the plane is $\vec{n} = \hat{i} + 2\hat{j} + 3\hat{k}$.The dot product is $\vec{b} \cdot \vec{n} = (3)(1) + (1)(2) + (0)(3) = 3 + 2 + 0 = 5$.The magnitudes are $|\vec{b}| = \sqrt{3^2 + 1^2} = \sqrt{10}$ and $|\vec{n}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}$.Substituting these values into the formula: $\sin 	heta = \frac{5}{\sqrt{10} \cdot \sqrt{14}} = \frac{5}{\sqrt{140}} = \frac{5}{2\sqrt{35}} = \frac{\sqrt{5} \cdot \sqrt{5}}{2\sqrt{7} \cdot \sqrt{5}} = \frac{\sqrt{5}}{2\sqrt{7}}$.Thus,$	heta = \sin^{-1}\left(\frac{\sqrt{5}}{2\sqrt{7}}ight)$.

The angle between the line $\vec{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(3\hat{i}+\hat{j})$ and the plane $\vec{r} \cdot (\hat{i}+2\hat{j}+3\hat{k})=8$ is:

Similar Questions

The foot of the perpendicular drawn from the point $(1, 1, 1)$ to the plane $\pi_1$ is $(1, 3, 5)$. If $(2, 2, -1), (3, 4, 2), (3, 3, 0)$ are three points on the plane $\pi_2$, then the angle between the planes $\pi_1$ and $\pi_2$ is

$A$ plane $E$ is perpendicular to the two planes $2x - 2y + z = 0$ and $x - y + 2z = 4$,and passes through the point $P(1, -1, 1)$. If the distance of the plane $E$ from the point $Q(a, a, 2)$ is $3\sqrt{2}$,then $(PQ)^2$ is equal to

What is the point of intersection of the line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z + 2}{3}$ and the plane $2x + 3y + z = 0$?

Find the ratio in which the plane $x - 2y + 3z = 17$ divides the line segment joining the points $(-2, 4, 7)$ and $(3, -5, 8)$.

The equation of the line passing through the point $(2, 3, 1)$ and parallel to the line of intersection of the planes $x - 2y - z + 5 = 0$ and $x + y + 3z = 6$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module