Question

Find the distance between the point (7, 2, 4) and the plane determined by the points A(2, 5, −3), B(−2, −3, 5) and C(5, 3, −3). [CBSE 2014]

Answer 1

The given points are A(2, 5, −3), B(−2, −3, 5) and C(5, 3, −3).

The equation of the plane ABC is given by

$$\begin{gathered} \left|\begin{array}{ccc}x-x_1& y-y_1& z-z_1\\ x_2-x_1& y_2-y_1& z_2-z_1\\ x_3-x_1& y_3-y_1& z_3-z_1\end{array}\right|=0 \\ \Rightarrow \left|\begin{array}{ccc}x-2& y-5& z-\left(-3\right)\\ -2-2& -3-5& 5-\left(-3\right)\\ 5-2& 3-5& -3-\left(-3\right)\end{array}\right|=0 \\ \Rightarrow \left|\begin{array}{ccc}x-2& y-5& z+3\\ -4& -8& 8\\ 3& -2& 0\end{array}\right|=0 \\ \Rightarrow \left|\begin{array}{ccc}x-2& y-5& z+3\\ 1& 2& -2\\ 3& -2& 0\end{array}\right|=0 \\ \Rightarrow -4\left(x-2\right)-6\left(y-5\right)-8\left(z+3\right)=0 \\ \Rightarrow 2\left(x-2\right)+3\left(y-5\right)+4\left(z+3\right)=0 \\ \Rightarrow 2x+3y+4z-7=0 \end{gathered}$$

∴ Distance between the point (7, 2, 4) and the plane $2x+3y+4z-7=0$

= Length of perpendicular from (7, 2, 4) to the plane $2x+3y+4z-7=0$

$$\begin{gathered} =\left|\frac{2\times 7+3\times 2+4\times 4-7}{\sqrt{2^2+3^2+4^2}}\right| \\ =\left|\frac{14+6+16-7}{\sqrt{4+9+16}}\right| \\ =\left|\frac{29}{\sqrt{29}}\right| \\ =\sqrt{29}\mathrm{units} \end{gathered}$$

Thus, the required distance between the given point and the plane is $\sqrt{29}$ units.