The angle between the lines vec{r}=(2 hat{i}- | vedclass

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

Question

(A) The given lines are in the form $\vec{r} = \vec{a} + \lambda \vec{b}$.For the first line,the direction vector is $\vec{b_1} = \hat{i} + 4 \hat{j}

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(A) The given lines are in the form $\vec{r} = \vec{a} + \lambda \vec{b}$.For the first line,the direction vector is $\vec{b_1} = \hat{i} + 4 \hat{j} + 3 \hat{k}$.For the second line,the direction vector is $\vec{b_2} = \hat{i} + 2 \hat{j} - 3 \hat{k}$.The angle $	heta$ between the two lines is given by $\cos 	heta = \frac{|\vec{b_1} \cdot \vec{b_2}|}{|\vec{b_1}| |\vec{b_2}|}$.Calculating the dot product: $\vec{b_1} \cdot \vec{b_2} = (1)(1) + (4)(2) + (3)(-3) = 1 + 8 - 9 = 0$.Since the dot product is $0$,the angle between the lines is $	heta = \cos^{-1}(0) = \frac{\pi}{2}$.

The angle between the lines $\vec{r}=(2 \hat{i}-3 \hat{j}+\hat{k})+\lambda(\hat{i}+4 \hat{j}+3 \hat{k})$ and $\vec{r}=(\hat{i}-\hat{j}+2 \hat{k})+\mu(\hat{i}+2 \hat{j}-3 \hat{k})$ is

Similar Questions

The equation of the straight line passing through the points $(4, -5, -2)$ and $(-1, 5, 3)$ is

$A(-1, 2, -3), B(5, 0, -6), C(0, 4, -1)$ are the vertices of a triangle $ABC$. The direction cosines of the internal bisector of $\angle BAC$ are

$\triangle ABC$ 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 angle between the two lines $\frac{x-2}{2} = \frac{2-y}{3} = \frac{z-1}{2}$ and $\frac{x-1}{2} = \frac{y+1}{1} = \frac{z-3}{-3}$ is . . . . . . .

The length of the perpendicular drawn from the point $(3, -1, 11)$ to the line $\frac{x}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module