Question

The value of $\lambda$ for which the triangle $ABC$ whose vertices are $A(6,10,10),B(1,0,-5)$ and $C(6,-10,\lambda )$ is right-angled at $B$,is

• A. $0$
• B. $30$
• C. $\frac{70}{3}$
• D. $\frac{3}{70}$

Answer 1

The correct option is A $0$

Given $A(6,10,10)B(1,0,-5)C(6,-10,\lambda )$ are vertices of triangle $ABC$ right angle at $B$

Formula: Vector equation is if points are $(x_1,y_1,z_1)(x_2,y_2,z_2)$ is $(x_1-x_2)\hat{i}+(y_1-y_2)\hat{j}+(z_1-z_2)\hat{k}$

Vector equation of $AB$ is $5\hat{i}+10\hat{j}+15\hat{k}$

Vector equation of $BC$ is $-5\hat{i}+10\hat{j}+(-5-\lambda )\hat{k}$

Given angle between $AB$ and $BC$ is ${90}^o$ then dot product is $0$

$-25+100+15(-5-\lambda )=015(-5-\lambda )=-\frac{75}{5}\lambda =0$

Value of $\lambda =0$
Option: $A$