Question

Let A, B, C be points with position vectors

r₁ = 2 $\hat { \mathbf { i } }$ – + , r₂ = $\hat { \mathbf { i } }$ +2+ 3 and r₃ = 3 $\hat { \mathbf { i } }$ ++ 2 relative to the origin 'O'. The shortest distance between point B and plane OAC is

• A. 10
• B. 5
• C. $\sqrt { \frac { 5 } { 7 } }$
• D. 2 $\sqrt { \frac { 5 } { 7 } }$

Answer 1

Answer: (D)

2 $\sqrt { \frac { 5 } { 7 } }$

Answer 2

Shortest distance between B and plane OAC

(h) =

here $\overrightarrow { \mathrm { OA } }$× $\overrightarrow { \mathrm { OC } }$ . $\overrightarrow { \mathrm { OB } }$= $\left| \begin{array} { c c c } 1 & 2 & 3 \\ 2 & - 1 & 1 \\ 3 & 1 & 2 \end{array} \right|$

= 1(–2 –1) + 2(3 – 4) + 3 (2 + 3) = 10

$\overrightarrow { \mathrm { OA } }$× $\overrightarrow { \mathrm { OC } }$ = $\left| \begin{array} { c c c } \hat { \mathrm { i } } & \hat { \mathrm { j } } & \hat { \mathrm { k } } \\ 2 & - 1 & 1 \\ 3 & 1 & 2 \end{array} \right|$ = $\hat { \mathbf { i } }$ (–2– 1) + (3 –4) + (2 + 3)

= –3 $\hat { \mathbf { i } }$ – + 5

|$\overrightarrow { \mathrm { OA } }$× $\overrightarrow { \mathrm { OC } }$ | = $\sqrt { 35 }$

h = $\frac { 10 } { \sqrt { 35 } }$= 2 $\sqrt { \frac { 5 } { 7 } }$