The acute angle between the line joining the | vedclass

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

Question

(C) The direction ratios (d.r.s.) of the line joining the points $P(2,1,-3)$ and $Q(-3,1,7)$ are given by $(x_2-x_1, y_2-y_1, z_2-z_1) = (-3-2, 1-1, 7

Vedclass · Accepted Answer

$\cos ^{-1}\left(\frac{7}{5 \sqrt{10}}\right)$

Vedclass · Answer

(C) The direction ratios (d.r.s.) of the line joining the points $P(2,1,-3)$ and $Q(-3,1,7)$ are given by $(x_2-x_1, y_2-y_1, z_2-z_1) = (-3-2, 1-1, 7-(-3)) = (-5, 0, 10)$.The direction ratios of the line parallel to $\frac{x-1}{3}=\frac{y}{4}=\frac{z+3}{5}$ are $(3, 4, 5)$.Let $	heta$ be the acute angle between these two lines. The formula for the angle between two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ is $\cos 	heta = \left| \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} ight|$.Substituting the values:$\cos 	heta = \left| \frac{(-5)(3) + (0)(4) + (10)(5)}{\sqrt{(-5)^2 + 0^2 + 10^2} \sqrt{3^2 + 4^2 + 5^2}} ight|$$\cos 	heta = \left| \frac{-15 + 0 + 50}{\sqrt{25 + 100} \sqrt{9 + 16 + 25}} ight|$$\cos 	heta = \frac{35}{\sqrt{125} \sqrt{50}} = \frac{35}{5\sqrt{5} \cdot 5\sqrt{2}} = \frac{35}{25\sqrt{10}} = \frac{7}{5\sqrt{10}}$.Therefore,$	heta = \cos ^{-1}\left(\frac{7}{5 \sqrt{10}}ight)$.

The acute angle between the line joining the points $(2,1,-3)$ and $(-3,1,7)$ and a line parallel to $\frac{x-1}{3}=\frac{y}{4}=\frac{z+3}{5}$ is

Similar Questions

Let a line pass through two distinct points $P(-2, -1, 3)$ and $Q$,and be parallel to the vector $3\hat{i} + 2\hat{j} + 2\hat{k}$. If the distance of the point $Q$ from the point $R(1, 3, 3)$ is $5$,then the square of the area of $\triangle PQR$ is equal to:

The shortest distance between the lines $\vec{r} = (\frac{1}{3}\hat{i} + 2\hat{j} + \frac{8}{3}\hat{k}) + \lambda(2\hat{i} - 5\hat{j} + 6\hat{k})$ and $\vec{r} = (-\frac{2}{3}\hat{i} - \frac{1}{3}\hat{k}) + \mu(\hat{i} - \hat{k})$, where $\lambda, \mu \in R$, is:

If $P$ is a point on the line parallel to the vector $2 \hat{i}-3 \hat{j}-6 \hat{k}$ and passing through the point $A$ whose position vector is $\hat{i}+2 \hat{j}-2 \hat{k}$ and $AP=21$, then the position vector of $P$ can be

If lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z-0}{1}$ intersect,then the value of $k$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module