The triad $(x, y, z)$ of real number such that $(3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})=(2…
The triad $(x, y, z)$ of real number such that $(3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})=(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}}) x+$ $\quad(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}) y+(-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}) z$ is
(–2, 5, 3)
(2, –5, 3)
(2, 5, 3)
(2, 5, –3)
Solution
It is given that,
$
\begin{array}{r}
(3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})=(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}}) x+(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \\
y+(-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}) z \\
=\hat{\mathbf{i}}(2 x+y-2 z)+\hat{\mathbf{j}}(3 x-2 y+z)+\hat{\mathbf{k}}(-x+2 y-2 z)
\end{array}
$
On comparing both the sides, we get
$
\begin{aligned}
& 2 x+y-2 z=3 \\
& 3 x-2 y+z=-1 \\
& -x+2 y-2 z=2
\end{aligned}
$
Now, by checking the options for coordinates $(\mathrm{x}, \mathrm{y}, \mathrm{z})$, option (c) satisfies them