Question

A plane passes through (2,3,-1) and is perpendicular to the line having direction ratios 3, -4, 7. The perpendicular distance from the origin to the plane is:
(a) $\dfrac{13}{\sqrt{74}}$
(b) $\dfrac{5}{\sqrt{74}}$
(c) $\dfrac{6}{\sqrt{74}}$
(d) $\dfrac{14}{\sqrt{74}}$

Answer 1

Hint: We know that for a general plane ax+by+cz+d=0, $a\widehat{i}+b\widehat{j}+c\widehat{k}$ represents the vector normal to the given plane, so the DRs of the line are the a, b and c of the plane. So, d is still unknown which can be found by putting the point (2,3,-1) in the equation of the plane. Once you get the equation of the plane, use the formula $\dfrac{d}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$ to get the distance of origin from the plane.

Complete step-by-step answer:
Let us start the solution to the above question. We know that for a general plane ax+by+cz+d=0, $a\widehat{i}+b\widehat{j}+c\widehat{k}$ represents the vector normal to the given plane. Also, it is given that the line with direction ratios 3, -4, 7 is perpendicular to the plane, so the a, b and c of this plane is 3, -4 and 7 respectively. So, the equation of the plane is:
\[ax+by+cz+d=0\]
\[\Rightarrow 3x-4y+7z+d=0\]
Now it is given that (2,3,-1) lies on the plane, so must satisfy the equation of the plane. If we put it in the equation of the plane we get:
\[3x-4y+7z+d=0\]
\[\Rightarrow 3\times 2-4\times 3+7\times \left( -1 \right)+d=0\]
\[\Rightarrow d=13\]
So, the equation of the plane is: \[3x-4y+7z+13=0\]
Now we know that the distance of a plane from the origin is $\dfrac{d}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$ .
$\dfrac{d}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}=\dfrac{13}{\sqrt{{{3}^{2}}+{{4}^{2}}+{{7}^{2}}}}=\dfrac{13}{\sqrt{74}}$
Therefore, the answer to the above question is option (a).

Note: In questions related to lines and planes the key thing is to remember the important formulas. It is also important that you know that the perpendicular distance of the point $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ is given by $\dfrac{a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$ .