Question

lf $AB\perp BC$, then the value of $\lambda$ equal, where $A(2k,2,3),B(k,1,5),C(3+k,2,1)$

• A. $3$
• B. $\frac{1}{3}$
• C. $-3$
• D. $-\frac{1}{3}$

Answer 1

The correct option is C $-3$

The dr's of $AB$ are $(k,1,-2)$
The dr's of $BC$ are $(3,1,-4)$
Since, they are perpendicular, $AB.BC=0$
$\Rightarrow 3k+1+8=0$
$\Rightarrow k=-3$