If theta is the angle between the lines frac{ | vedclass

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

Question

(B) Step $1$: Rewrite the equations of the lines in standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.For the first line: $\frac{x-1

Vedclass · Accepted Answer

$	heta = \frac{\pi}{3}$

Vedclass · Answer

(B) Step $1$: Rewrite the equations of the lines in standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.For the first line: $\frac{x-1}{2} = \frac{y - (-3/2)}{2} = \frac{z - (-2)}{0}$. The direction vector is $\vec{v_1} = 2\hat{i} + 2\hat{j} + 0\hat{k}$.Step $2$: For the second line: $x=1$ implies the direction ratio for $x$ is $0$. Thus, $\frac{x-1}{0} = \frac{y-1}{2} = \frac{z+1}{2}$. The direction vector is $\vec{v_2} = 0\hat{i} + 2\hat{j} + 2\hat{k}$.Step $3$: The cosine of the angle $	heta$ between the lines is given by $\cos 	heta = \frac{|\vec{v_1} \cdot \vec{v_2}|}{|\vec{v_1}| |\vec{v_2}|}$.Step $4$: Calculate the dot product: $\vec{v_1} \cdot \vec{v_2} = (2)(0) + (2)(2) + (0)(2) = 4$.Step $5$: Calculate the magnitudes: $|\vec{v_1}| = \sqrt{2^2 + 2^2 + 0^2} = \sqrt{8} = 2\sqrt{2}$ and $|\vec{v_2}| = \sqrt{0^2 + 2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$.Step $6$: $\cos 	heta = \frac{4}{(2\sqrt{2})(2\sqrt{2})} = \frac{4}{8} = \frac{1}{2}$.Step $7$: Therefore, $	heta = \cos^{-1}(\frac{1}{2}) = \frac{\pi}{3}$.

If $\theta$ is the angle between the lines $\frac{x - 1}{2} = \frac{2y + 3}{4}; z = -2$ and $x = 1; \frac{y - 1}{2} = \frac{z + 1}{2}$, then

Similar Questions

$\triangle ABC$ is formed by $A(1, 8, 4)$,$B(0, -11, 4)$,and $C(2, -3, 1)$. If $D$ is the foot of the perpendicular drawn from $A$ to $BC$,then the coordinates of $D$ are

The measure of the angle between the lines $x = k + 1, y = 2k - 1, z = 2k + 3, k \in R$ and $\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-3}{1}$ is

The length of the perpendicular drawn from the point $2 \hat{i} - \hat{j} + 5 \hat{k}$ to the line $\vec{r} = (11 \hat{i} - 2 \hat{j} - 8 \hat{k}) + \lambda(10 \hat{i} - 4 \hat{j} - 11 \hat{k})$ is

Find the coordinates of the foot of the perpendicular drawn from the point $A(1, 8, 4)$ to the line joining the points $B(0, -1, 3)$ and $C(2, -3, -1)$.

If the lines $\frac{x-1}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}$ and $\frac{x-1}{3k}=\frac{y-1}{1}=\frac{z-6}{-5}$ are perpendicular,find the value of $k$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module