Mathematics › Vectors › Product of 2 vectors
If $\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+8 \hat{\mathrm{k}},…
If $\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+8 \hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=7 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$ are two vectors and $(2 \vec{a}-3 \vec{b}) \times(4 \vec{a}+\vec{b})=x \hat{i}+y \hat{j}+z \hat{k}$, then $x+y+z=$
-1000 1400 1000 -1400
Solution
$\begin{aligned} & \text { } \overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+8 \hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=7 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} \\ & (2 \overrightarrow{\mathrm{a}}-3 \overrightarrow{\mathrm{b}}) \times(4 \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})=x \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}\end{aligned}$
$
\begin{aligned}
& 2 \vec{a}-3 \vec{b}=-17 \hat{i}+5 \hat{j}+7 \hat{k} \\
& 4 \vec{a}+\vec{b}=15 \hat{i}-25 \hat{j}+35 \hat{k}
\end{aligned}
$
Now, $(2 \overrightarrow{\mathrm{a}}-3 \overrightarrow{\mathrm{b}}) \times(4 \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ -17 & 5 & 7 \\ 15 & -25 & 35\end{array}\right|$
$
\begin{aligned}
& =\hat{\mathrm{i}}(175+175)-\hat{\mathrm{j}}(-595-105)+\hat{\mathrm{k}}(25 \times 17-75) \\
& \Rightarrow(2 \overrightarrow{\mathrm{a}}-3 \overrightarrow{\mathrm{b}}) \times(4 \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})=350 \hat{\mathrm{i}}+700 \hat{\mathrm{j}}+350 \hat{\mathrm{k}}
\end{aligned}
$
From eqns. (i) \& (ii)
$
\begin{aligned}
& x=350, y=700 \& z=350 \\
& \therefore x+y+z=350+700+350=140
\end{aligned}
$
Asked in: AP EAMCET 2023 (18 May Shift 2)
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