MCQ
If three points $A, B$ and $C$ have position vectors $\hat{\text{i}}+\text{x}\hat{\text{j}}+3\hat{\text{k}},\ 3\hat{\text{i}}+4\hat{\text{j}}+7\hat{\text{k}}$ and $\text{y}\hat{\text{i}}-2\hat{\text{j}}-5\hat{\text{k}}$ respectively are collinear, then $(x, y) =$
  • $(2, -3)$
  • B
    $(-2, 3)$
  • C
    $(-2, -3)$
  • D
    $(2, 3)$

Answer

Correct option: A.
$(2, -3)$
Given position vector of $A, B$ and $C$ are $\hat{\text{i}}+\text{x}\hat{\text{j}}+3\hat{\text{k}},\ 3\hat{\text{i}}+4\hat{\text{j}}+7\hat{\text{k}}$ and $\text{y}\hat{\text{i}}-2\hat{\text{j}}-5\hat{\text{k}}$. Then,
$\overrightarrow{\text{AB}}=3\hat{\text{i}}+4\hat{\text{j}}+7\hat{\text{k}}-\hat{\text{i}}-\text{x}\hat{\text{j}}-3\hat{\text{k}}$
$=2\hat{\text{i}}+(4-\text{x})\hat{\text{j}}+4\hat{\text{k}}$
$\overrightarrow{\text{BC}}=\text{y}\hat{\text{i}}-2\hat{\text{j}}-5\hat{\text{k}}-3\hat{\text{i}}-4\hat{\text{j}}-7\hat{\text{k}}$
$=(\text{y}-3)\hat{\text{i}}-6\hat{\text{j}}-12\hat{\text{k}}$
Since, the given vectors are collinear.
$\therefore\ \overrightarrow{\text{AB}}=\lambda\overrightarrow{\text{BC}}$
$\Rightarrow\ 2\hat{\text{i}}+(4-\text{x})\hat{\text{j}}+4\hat{\text{k}}=\lambda(\text{y}-3)\hat{\text{i}}-6\lambda\hat{\text{j}}-12\lambda\hat{\text{k}}$
$\Rightarrow\ 2=\lambda(\text{y}-3),\ (4-\text{x})=-6\lambda,\ 4=-12\lambda$
$\Rightarrow\ 2=\lambda(\text{y}-3),\ (4-\text{x})=-6\lambda,\ \lambda=-\frac{1}3$
$\Rightarrow\ 2=-\frac{1}3(\text{y}-3),\ (4-\text{x})=-6\times\Big(-\frac{1}3\Big)$
$\Rightarrow\ -6=\text{y}-3,\ 4-\text{x}=2$
$\Rightarrow\ \text{y}=-3,\ \text{x}=2$

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