Let P be the plane passing through the points (5,3,0),(13,3,-2) and (1,6,2).

Question

Let P be the plane passing through the points $(5,3,0),(13,3,-2)$ and $(1,6,2)$.

For $\alpha \in N$, if the distances of the points $A(3,4,\alpha )$ and $B(2,\alpha ,a)$ from the plane P are 2 and 3 respectively, then the positive value of a is :

Accepted Answer

$$
\begin{aligned}
& \overrightarrow{A B}=8 \hat{i}-2 \hat{k} \\
& \overrightarrow{A C}=-4 \hat{i}+3 \hat{j}+2 \hat{k}
\end{aligned}
$$

$$
\begin{aligned}
& \overrightarrow{A B} \times \overrightarrow{A C}=\left|\[\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \ 8 & 0 & -2 \ -4 & 3 & 2 \end{array}\end{aligned}$$\right| \\
\& =6 \hat{i}-8 \hat{j}+24 \hat{k}

\]

Equation of plane : $6 x-8 y+24 z=d$ passes through $(5,3,0)$

$$
\begin{aligned}
& 6 \times 5-8 \times 3+24 \times 0=d \\
& \Rightarrow d=6 \\
& 6 x-8 y+24 z=6 \\
& \Rightarrow 3 x-4 y+12 z=3
\end{aligned}
$$

Distance of point $(3,4, \alpha)$

$$
\frac{9-16+12 \alpha-3}{\sqrt{9+16+144}}=2 \Rightarrow \alpha=3
$$

Distance of point $(2, \alpha, a)$

$$
\begin{aligned}
& \frac{3 \times 2-4 \times 3+12 \times a-3}{13}=3 \\
& \Rightarrow 12 a-9=39 \\
& \Rightarrow 12 a=48 \\
& \Rightarrow a=4
\end{aligned}
$$

Solutions · 1