Question

If M denotes the mid-point of the line joining

A(4$\widehat{i}$ + 5$\widehat{j}$ –10$\widehat{k}$) & B(–$\widehat{i}$ + 2$\widehat{j}$ + $\widehat{k}$), then the equation of the plane through M & perpendicular to AB is :

• A. $\overset{\rightarrow}{r}$.(–5$\widehat{i}$ – 3$\widehat{j}$ +11$\widehat{k}$)+$\frac{135}{2}$ = 0
• B. $\overset{\rightarrow}{r}$.$\left( \frac{3}{2}\widehat{i} + \frac{7}{2}\widehat{j}–\frac{9}{2}\widehat{k} \right)$ + $\frac{135}{2}$ = 0
• C. $\overset{\rightarrow}{r}$. (4$\widehat{i}$ + 5$\widehat{j}$ –10$\widehat{k}$) + 4 = 0
• D. $\overset{\rightarrow}{r}$. (–$\widehat{i}$ + 2$\widehat{j}$+ $\widehat{k}$) + 4 = 0

Answer 1

Answer: (A)

$\overset{\rightarrow}{r}$.(–5$\widehat{i}$ – 3$\widehat{j}$ +11$\widehat{k}$)+$\frac{135}{2}$ = 0

Answer 2

Q M is the mid point of AB

\ M =$\left\{ \frac{(4\widehat{i} + 5\widehat{j}–10\widehat{k}) + (–\widehat{i} + 2\widehat{j} + \widehat{k})}{2} \right\}$

= $\frac{1}{2}$ (3$\widehat{i}$ + 7$\widehat{j}$ – 9$\widehat{k}$) = $\overset{\rightarrow}{a}$

$\overset{\rightarrow}{n}$ = $\overset{\rightarrow}{AB}$ = normal vector of the plane

Q plane is perpendicular to the line AB

Ž $\overset{\rightarrow}{n}$ = $\overset{\rightarrow}{AB}$ = $\overset{\rightarrow}{OB}$– $\overset{\rightarrow}{OA}$ = (–5$\widehat{i}$ – 3$\widehat{j}$ + 11$\widehat{k}$)

\ Equation of plane is $\overset{\rightarrow}{r}$.$\overset{\rightarrow}{n}$ = $\overset{\rightarrow}{a}$. $\overset{\rightarrow}{n}$

where $\overset{\rightarrow}{a}$ = $\overset{\rightarrow}{OM}$