Question
Using vector method, prove that the point is collinear:
A(1, 2, 7), B(2, 6, 3) and C(3, 10, -1)
A(1, 2, 7), B(2, 6, 3) and C(3, 10, -1)
$=2\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}}-\hat{\text{i}}-2\hat{\text{j}}-7\hat{\text{k}}$
$=\hat{\text{i}}+4\hat{\text{j}}-4\hat{\text{k}}$
$\overrightarrow{\text{BC}}=$
Position vector of C - Position vector of B$=3\hat{\text{i}}+10\hat{\text{j}}-\hat{\text{k}}-2\hat{\text{i}}-6\hat{\text{j}}-3\hat{\text{k}}$
$=\hat{\text{i}}+4\hat{\text{j}}-4\hat{\text{k}}$
$\therefore\ \overrightarrow{\text{AB}}=\overrightarrow{\text{BC}}$
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.
$\frac{\text{a}}{\sin\text{A}}=\frac{\text{b}}{\sin\text{B}}=\frac{\text{c}}{\sin\text{C}}$
Where a, b and c are lengths of the sides opposite, respectively, to the angles A, B and C of $\Delta$ ABC.$\int\text{x}^2\text{e}^{-\text{x}}\text{dx}$