The distance of the point A ( - 2 ,   3 ,   1 ) from the line through P ( - 3 ,   5 ,  …

1. $\frac{2}{\sqrt{3}}$
2. $\sqrt{\frac{14}{3}}$
3. $\frac{16}{\sqrt{3}}$
4. $\frac{5}{\sqrt{3}}$

Solution

Here, $\alpha =\beta =\gamma$

$\because {\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =1$

$\therefore \cos \alpha =\frac{1}{\sqrt{3}}$

DC' of $PQ$ are $\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$




$PM=$ Projection of $AP$ on $PQ$

$=\left|\left(-2+3\right)\frac{1}{\sqrt{3}}+\left(3-5\right)\cdot \frac{1}{\sqrt{3}}+\left(1-2\right)\cdot \frac{1}{\sqrt{3}}\right|=\frac{2}{\sqrt{3}}$

And $AP=\sqrt{{\left(-2+3\right)}^2+{\left(3-5\right)}^2+{\left(1-2\right)}^2}=\sqrt{6}$

$AM=\sqrt{{\left(AP\right)}^2-{\left(PM\right)}^2}$

$=\sqrt{6-\frac{4}{3}}=\sqrt{\frac{14}{3}}$

Asked in: MHT CET Full Test 2

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