The angle between the pair of lines frac{x+3} | vedclass

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

Question

(D) The direction ratios of the first line are $\vec{b_1} = \langle 3, 5, 4 \rangle$. The direction ratios of the second line are $\vec{b_2} = \langle

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) The direction ratios of the first line are $\vec{b_1} = \langle 3, 5, 4 angle$. The direction ratios of the second line are $\vec{b_2} = \langle 1, 4, 2 angle$. The angle $	heta$ between two lines with direction ratios $\langle a_1, b_1, c_1 angle$ and $\langle a_2, b_2, c_2 angle$ is given by $\cos 	heta = \left| \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} ight|$. Substituting the values: $\cos 	heta = \left| \frac{(3)(1) + (5)(4) + (4)(2)}{\sqrt{3^2 + 5^2 + 4^2} \sqrt{1^2 + 4^2 + 2^2}} ight|$ $\cos 	heta = \left| \frac{3 + 20 + 8}{\sqrt{9 + 25 + 16} \sqrt{1 + 16 + 4}} ight|$ $\cos 	heta = \left| \frac{31}{\sqrt{50} \sqrt{21}} ight| = \frac{31}{5 \sqrt{2} \sqrt{21}} = \frac{31}{5 \sqrt{42}}$. Thus,$	heta = \cos^{-1} \left( \frac{31}{5 \sqrt{42}} ight)$. Since this value is not present in the given options,the correct choice is $D$.

The angle between the pair of lines $\frac{x+3}{3}=\frac{y-1}{5}=\frac{z+3}{4}$ and $\frac{x+1}{1}=\frac{y-4}{4}=\frac{z-5}{2}$ is

Similar Questions

Find the shortest distance between the lines whose vector equations are $\vec{r}=(1-t) \hat{i}+(t-2) \hat{j}+(3-2 t) \hat{k}$ and $\vec{r}=(s+1) \hat{i}+(2 s-1) \hat{j}-(2 s+1) \hat{k}$.

If the shortest distance between the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{\lambda}$ and $\frac{x-2}{1}=\frac{y-4}{4}=\frac{z-5}{5}$ is $\frac{1}{\sqrt{3}}$,then the sum of possible values of $\lambda$ is

The point of intersection of the lines $\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}$ and $\frac{x - 2}{1} = \frac{y - 4}{3} = \frac{z - 6}{5}$ is

The perpendicular distance from the point $P(3, 5, 2)$ to the line $L$ passing through the point $2\hat{i} + \hat{j}$ and parallel to the vector $\hat{i} + 5\hat{j} + 2\hat{k}$ is

If the shortest distance between the lines $\frac{x-\lambda}{2}=\frac{y-4}{3}=\frac{z-3}{4}$ and $\frac{x-2}{4}=\frac{y-4}{6}=\frac{z-7}{8}$ is $\frac{13}{\sqrt{29}}$,then a value of $\lambda$ is :

Vedclass Test Series

Exam Paper Generator

Online Exam Module