$\vec{a}=4 \hat{i}+13 \hat{j}-18 \hat{k}, \vec{b}=\hat{i}-2 \hat{j}+3 \hat{k}$ 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 $x+y=$

Question

Aicharya · Accepted Answer

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

$\vec{a}=4 \hat{i}+13 \hat{j}-18 \hat{k}, \vec{b}=\hat{i}-2 \hat{j}+3 \hat{k}$ and $\vec{c}=2 \hat{i}+3…

Solution