Question

If a line $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ lies in a plane $a_{2}x+b_{2}y+c_{2}z=d,$ then which of the following is / are correct -

• A.
• B.
• C.
• D. None of these

Answer 1

The correct option is A

We know in the given equation of line $a_1,b_1,c_1$ are the direction ratios of the line and in the equation of plane $a_2,b_2,c_2$ are the direction ratios of the normal to the plane. So we have two vector with us. All we need now is the angle between them.
Let the angle between plane and line is $\theta$, then the angle between normal to the plane and line is ${90}^{\circ }-\theta$.
$\cos ({90}^{\circ }-\theta )=\frac{a.b}{|a||b|}$
$\cos ({90}^{\circ }-\theta )=\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{(a_1^2+b_1^2+c_1^2)(a_2^2+b_2^2+c_2^2)}}$
Or $\sin \left(\theta \right)=\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{(a_1^2+b_1^2+c_1^2)(a_2^2+b_2^2+c_2^2)}}$
If the line lies in the plane then the angle between line and the plane becomes $0^{\circ }$.
So, $\sin (0^{\circ }$) = $\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{(a_1^2+b_1^2+c_1^2)(a_2^2+b_2^2+c_2^2)}}$
Or $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$