Find the angle between the lines 2x=3y=-z and 6x=-y=-4z. - Vidyadip

Question

Find the angle between the lines $2\text{x}=3\text{y}=-\text{z}$ and $6\text{x}=-\text{y}=-4\text{z}.$

✓

Answer

The equations of the given lines can be re-writen as
$\frac{\text{x}}{3}=\frac{\text{y}}{2}=\frac{\text{z}}{-6}$ and $\frac{\text{x}}{2}=\frac{\text{y}}{-12}=\frac{\text{z}}{-3}$
We know that angle between the lines $\frac{\text{x}-\text{x}_1}{\text{a}_1}=\frac{\text{y}-\text{y}_1}{\text{b}_1}=\frac{\text{z}-\text{z}_1}{\text{c}_1}$ and $\frac{\text{x}-\text{x}_2}{\text{a}_2}=\frac{\text{y}-\text{y}_2}{\text{b}_2}=\frac{\text{z}-\text{z}_2}{\text{c}_2}$ is given by $\cos\theta=\frac{\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2}{\sqrt{\text{a}^2_1+\text{b}^2_1+\text{c}^2_1}\sqrt{\text{a}^2_2+\text{b}^2_2+\text{c}^2_2}}.$
Let $\theta$ be the angle between the given lines.
$\therefore\cos\theta=\frac{3\times2+2\times(-12)+(-6)\times(-3)}{\sqrt{3^2+2^2+(-6)^2}\sqrt{2^2+(-12)^2+(-3)^2}}$
$=\frac{6-24+18}{\sqrt{49}\sqrt{157}}$
$=0$
$\Rightarrow\theta=\frac{\pi}{2}$
Thus, the angle between the given lines is $\frac{\pi}{2}.$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "3 Marks Question" from Straight line in space→Straight line in space — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

In the following, determine the values of constants involved in the definition so that the given function is continuous:
$\text{f(x)}=\begin{cases}\text{k}(\text{x}^2+3\text{x}),&\text{if }\text{ x}<0\\\cos2\text{x},&\text{if }\text{ x}\geq0\end{cases}$

Find the perpendicular distence of the point (3, -1, 11) from the line $\frac{\text{x}}{2}=\frac{\text{y}-2}{-3}=\frac{\text{z}-3}{4}.$

If $\vec{\text{a}}=2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}},\vec{\text{b}}=\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}$ and $\vec{\text{c}}=\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}}.$ find $\lambda$ such that $\vec{\text{a}}$ is perpendicular to $\lambda\vec{\text{b}}+\vec{\text{c}}.$

If $\text{A}=\begin{bmatrix}3&1\\-1&2\end{bmatrix}$ and $\text{I}=\begin{bmatrix}1&0\\0&1\end{bmatrix},$ then find $\lambda$ so that $\text{A}^2 = 5\text{A} + \lambda\text{I}.$

Let A = [-1, 1]. Then, discuss whether the following functions from A to itself are one-one, onto or bijective:
g(x) = |x|

Find the position vector of a point R which divides the line joining the two points P and Q with position vectors $\overrightarrow{\text{OP}}=2\vec{\text{a}}+\vec{\text{b}}\text{ and }\overrightarrow{\text{OQ}}=\vec{\text{a}}-2\vec{\text{b}}$, respectively in the ratio 1 : 2 internally and externally.

In the following, find the value of the constant k so that the given function is continuous at the indicated point:
$\text{f(x)}=\begin{cases}\frac{\text{x}^2-25}{\text{x}-5},&\text{x}\neq5\\\text{k},&\text{x}=5\end{cases}\text{at x} =5$

Find the equation of the plane with intercept 3 on the y-axis and parallel to the ZOX plane.

For what value of k is the function
$\text{f}\text{(x)}=\begin{cases}\frac{\sin5\text{x}}{3\text{x}}, &\text{if}\text{ x}\neq0\\\text{k}, &\text{if}\text{ x}=0\end{cases}$ is continuous at x = 0?

A can hit a target 3 times in 6 shots, B : 2 times in 6 shots and C : 4 times in 4 shots. They fix a volley. What is the probability that at least 2 shots hit?