Question

A line passing through $A(1,2,3)$ and having direction ratios $(3,4,5)$ meets a plane $x+2y-3z=5$ at $B$, then distance $AB$ is equal to

• A. $\frac{9}{4}$
• B. $\frac{11}{4}$
• C. $\frac{13}{4}$
• D. None of these.

Answer 1

The correct option is A $\frac{9}{4}$

Let equation of line through A,

$\Rightarrow ax+by+cz+d=0$

DR's $\to$ $a,b,c=3,4,5$

$\Rightarrow 3x+4y+5z+d=0$

Passes through $(1,2,3)$,

$\Rightarrow$$3+8+15+d=0$

$\Rightarrow$$d=-26$

Line equation is $3x+4y+5z-5=0⟶\left(1\right)$

$\Rightarrow x+2y-3z-5=0⟶\left(2\right)$

$\Rightarrow 3x+4y+5z-5=0⟶\left(2\right)$

We get B co-ordinates$(x_2,y_2,z_2).$

$A(1,2,3)\to (x_1,y_1,z_1)$

$AB=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2+{(z_2-z_2)}^2}$ $=\frac{9}{4}$