Three Dimensional Geometry — MATHS STD 12 Science — Question
Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSThree Dimensional Geometry5 Marks
Question
Find the angle between the lines whose direction cosines are given by the equations $l + m + n = 0, l^2 + m^2- n^2 = 0.$
✓
Answer
Eliminating $n$ from both the equations, we have $l^2 + m^2 - (l - m)^2 = 0$
$\Rightarrow l^2 + m^2 - l^2 - m^2 + 2ml = 0 $
$\Rightarrow 2lm = 0 $
$\Rightarrow lm = 0$
$\Rightarrow (-m - n)m = 0 [\because\text{l}=-\text{m}-\text{n}]$
$\Rightarrow m = -n$
$\Rightarrow m = 10$
$\Rightarrow l = 0, l = -n$
Thus, Dr’s two lines are proportional to $0, -n, n$ and $-n, 0, -1$ i.e., $0, -1, 1$ and $-1, 0, 1.$
So, the vector parallel to these given lines are $\vec{\text{a}}=-\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=-\hat{\text{i}}+\hat{\text{k}}$
Now, $\cos\theta=\frac{\vec{\text{a}}\vec{\text{b}}}{|\vec{\text{a}}||\vec{\text{b}}|}$
$\Rightarrow\frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}}$
$\Rightarrow\cos\theta=\frac{1}{2}$
$\therefore\theta=\frac{\pi}{3}\Big[\because\cos\frac{\pi}{3}=\frac{1}{2}\Big]$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.