Question

The triangle $ABC$ is defined by the vertices $A=(0,7,10)$ , $B=(-1,6,6)$ and $C=(-4,9,6)$. Let $D$ be the foot of the attitude from $B$ to the side $AC$ then $BD$ is

• A. $\overline{i}+2\overline{j}+2\overline{k}$
• B. $-\overline{i}+2\overline{j}+2\overline{k}$
• C. $\overline{i}+2\overline{j}-2\overline{k}$
• D. $\overline{i}-2\overline{j}+2\overline{k}$

Answer 1

Answer: (B)

$-\overline{i}+2\overline{j}+2\overline{k}$

Given : $A=(0,7,10),B=(-1,6,6)$ and $C=(-4,9,6)$

Position vector(P.V.) of $AB$ is $(-1,-1,-4)$

$⟹|AB|=\sqrt{(-1{)}^2+(-1{)}^2+(-4{)}^2}=\sqrt{18}$

P.V. of $BC$ is $(-3,3,0)$

$⟹|BC|=\sqrt{(-3{)}^2+3^2+0}=\sqrt{18}$

P.V. of $AC$ is $(-4,2,-4)$

$⟹|AC|=\sqrt{(-4{)}^2+2^2+(-4{)}^2}=6$

Now, $D$ is the midpoint of $AC$
$⟹D=(\frac{0-4}{2},\frac{7+9}{2},\frac{10+6}{2})=(-2,8,8)$
$⟹\overrightarrow{BD}=-\hat{i}+2\hat{j}+2\hat{k}$