Question

If the distance of the point P(4, 3, 5) from the Y-axis is $\lambda$, then the value of ${7\lambda }^2$ is

• A. $287$
• B. $7\sqrt{41}$
• C. $63$
• D. $21$

Answer 1

The correct option is A $287$

$\Rightarrow$ $A=(x_1,y_1,z_1)=(0,3,0)$ and $P=(x_2,y_2,z_2)=(4,3,5)$

Here, $PA=\lambda$

$\Rightarrow$ $AP\left(\lambda \right)=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$

$=\sqrt{(4-0{)}^2+(3-3{)}^2+(5-0{)}^2}$

$=\sqrt{16+0+25}$

$=\sqrt{41}$

$\Rightarrow$ $7{\lambda }^2=7(\sqrt{41}{)}^2$

$=7\times 41$

$=287$