Question
Using vector method, prove that the point is collinear:
A(6, -7, -1), B(2, -3, 1) and C(4, -5, 0)
A(6, -7, -1), B(2, -3, 1) and C(4, -5, 0)
$=2\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}}-6\hat{\text{i}}+7\hat{\text{j}}+\hat{\text{k}}$
$=-4\hat{\text{i}}+4\hat{\text{j}}+2\hat{\text{k}}$
$=-2\big(2\hat{\text{i}}-2\hat{\text{j}}-\hat{\text{k}}\big)$$\overrightarrow{\text{BC}}=$
Position vector of C - Position vector of B$=4\hat{\text{i}}-5\hat{\text{j}}-2\hat{\text{k}}+3\hat{\text{j}}-\hat{\text{k}}$
$=2\hat{\text{i}}-2\hat{\text{j}}-\hat{\text{k}}$
$\therefore\ \overrightarrow{\text{AB}}=-2\overrightarrow{\text{BC}}$
So, $\overrightarrow{\text{AB}}$ and $\overrightarrow{\text{BC}}$ are parallel vectors. But B is a point common to them. Hence, the given points A, B, and C are collinear.Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$(1-\text{y}^2)(1+\log\text{x})\text{dx}+2\text{xy dy}=0,$ given that $\text{y}=0$ when $\text{x}=1.$