Three Dimensional Geometry — Maths STD 12 Science — Question
Gujarat BoardEnglish MediumSTD 12 ScienceMathsThree Dimensional Geometry5 Marks
Question
Find the angle between the lines whose direction cosines are given by the equations $l + m + n = 0, l^2 + m^2- n^2 = 0.$
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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, $\cos\theta=\frac{\vec{\text{a}}\vec{\text{b}}}{|\vec{\text{a}}||\vec{\text{b}}|}\Rightarrow\frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}}$
$\Rightarrow\cos\theta=\frac{1}{2}$
$\therefore\theta=\frac{\pi}{3}\Big[\because\cos\frac{\pi}{3}=\frac{1}{2}\Big]$
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