$\mathbf{a}, \mathbf{b}, \mathbf{c}$ are non-coplanar vectors. If $\begin{aligned} & \mathbf{a}+3 \mathbf{b}+4 \mathbf{c}=x(\mathbf{a}-2 \mathbf{b}+3 \mathbf{c})+y(\mathbf{a}+5 \mathbf{b}-2 \mathbf{c}) \ & +z(6 \mathbf{a}+14 \mathbf{b}+4 \mathbf{c}), ext { then } x+y+z=\end{aligned}$

Question

$\mathbf{a}, \mathbf{b}, \mathbf{c}$ are non-coplanar vectors. If $\begin{aligned} & \mathbf{a}+3 \mathbf{b}+4 \mathbf{c}=x(\mathbf{a}-2 \mathbf{b}+3 \mathbf{c})+y(\mathbf{a}+5 \mathbf{b}-2 \mathbf{c}) \ & +z(6 \mathbf{a}+14 \mathbf{b}+4 \mathbf{c}), 	ext { then } x+y+z=\end{aligned}$

Aicharya · Accepted Answer

$\mathbf{a}+3 \mathbf{b}+4 \mathbf{c}=\mathbf{a}(x+y+6 z)+\mathbf{b}(-2 x+5 y+14 z)$ $+c(3 x-2 y+4 z)$ (because $a, b$ and $c$ are non-coplanar vectors) Comparing both sides, $x+y+6 z=1$ ...(i) $-2 x+5 y+14 z=3$ ...(ii) $3 x-2 y+4 z=4$ ...(iii) By solving Eqs. (i), (ii) and (iii), we get $\begin{aligned} & x=-2, y=-3, z=1 \ & x+y+z=-4\end{aligned}$

$\mathbf{a}, \mathbf{b}, \mathbf{c}$ are non-coplanar vectors. If $\begin{aligned} & \mathbf{a}+3…

Solution