Q54 / 150 · MHT CET 2021 (23 Sep Shift 1) · Mathematics

$\vec{a}=4 \hat{i}+13 \hat{j}-18 \hat{k}, \vec{b}=\hat{i}-2 \hat{j}+3 \hat{k}$ and $\vec{c}=2 \hat{i}+3 \hat{j}-4 \hat{k}$ are three vectors such that $\vec{a}=x \vec{b}+y \vec{c}$, then $x+y=$

  • A
    -1
  • B
    -2
  • C
    5
  • Correct answer
    D
    1

Solution

$$ \begin{aligned} & \vec{a}=x \vec{b}+y \vec{c} \\ & \therefore 4 \hat{i}+13 \hat{j}-18 \hat{k}=(\hat{i}-2 \hat{j}+3 \hat{k})(x)+(2 \hat{i}+3 \hat{j}-4 \hat{k})(y) \\ & =(x+2 y) \hat{i}+(-2 x+3 y) \hat{j}+(3 x-4 y) \hat{k} \\ & \therefore x+2 y=4,-2 x+3 y=13 \text { and } 3 x-4 y=-18 \end{aligned} $$ Solving, we get $x=-2, y=3 \Rightarrow x+y=1$

Hence, the correct answer is option: 'D'.

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