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Question: Answered & Verified by Expert
$\vec{a}=4 \hat{i}+13 \hat{j}-18 \hat{k}, \vec{b}=\hat{i}-2 \hat{j}+3 \hat{k}$ 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 $x+y=$
MathematicsVector AlgebraMHT CETMHT CET 2021 (23 Sep Shift 1)
Options:
  • A -1
  • B -2
  • C 5
  • D 1
Solution:
1421 Upvotes Verified Answer
The correct answer is: 1
$$
\begin{aligned}
& \vec{a}=x \vec{b}+y \vec{c} \\
& \therefore 4 \hat{i}+13 \hat{j}-18 \hat{k}=(\hat{i}-2 \hat{j}+3 \hat{k})(x)+(2 \hat{i}+3 \hat{j}-4 \hat{k})(y) \\
& =(x+2 y) \hat{i}+(-2 x+3 y) \hat{j}+(3 x-4 y) \hat{k} \\
& \therefore x+2 y=4,-2 x+3 y=13 \text { and } 3 x-4 y=-18
\end{aligned}
$$
Solving, we get $x=-2, y=3 \Rightarrow x+y=1$

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