Question

A plane meets coordinate axes at P , Q and R respectively such that position vector of centroid of $\Delta PQR$ is $(2\hat{i}-5\hat{j}+8\hat{k})$, then the equation of plane is

• A. $\overline{r}\cdot (20\hat{i}-\hat{j}+5\hat{k})=120$
• B. $\overline{r}\cdot (20\hat{i}-8\hat{j}+5\hat{k})=120$
• C. $\overline{r}\cdot (20\hat{i}-8\hat{j}+5\hat{k})=2$
• D. $\overline{r}\cdot (20\hat{i}-8\hat{j}+5\hat{k})=20$

Answer 1

Answer: (B)

$\overline{r}\cdot (20\hat{i}-8\hat{j}+5\hat{k})=120$

Let P ( x-axis) $=(p,0,0)$

Q ( y-axis) $=(0,q,0)$

R (z-axis) $=(0,0,r)$

Centroid $2,-5,8$

$\frac{p}{3}=2;p=6$

$\frac{q}{3}=-5;q=-15$

$\frac{r}{3}=8;r=24$

Intercept eqn of plane;

$\frac{x}{6}-\frac{y}{15}+\frac{z}{24}=1$

$\Rightarrow 20x-8y+5z=120$

$\Rightarrow \overrightarrow{r}.(20\hat{i}-8\hat{j}+5\hat{k})=120$