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$\mathbf{a}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}-\hat{\mathbf{j}}$ and $\mathbf{c}=5 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$, then unit vector parallel to $\mathbf{a}+\mathbf{b}-\mathbf{c}$ but in opposite direction is
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Verified Answer
The correct answer is:
$\frac{1}{3}(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})$
Given, $\mathbf{a}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}-\hat{\mathbf{j}}$ and $\mathbf{c}=5 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$
$\begin{aligned}
: \mathbf{a}+\mathbf{b}-\mathbf{c} & =(2+1-5) \hat{\mathbf{i}}+(1-1+1) \hat{\mathbf{j}}+(-1+0-1) \hat{\mathbf{k}} \\
& =-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}=-(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})
\end{aligned}$
Now, the unit vector in the direction of $\mathbf{a}+\mathbf{b}-\mathbf{c}$ be
$\frac{-(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})}{\sqrt{2^2+(-1)^2+2^2}}=-\frac{1}{3}(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})$
The required unit vector be $\frac{1}{3}(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})$.
$\begin{aligned}
: \mathbf{a}+\mathbf{b}-\mathbf{c} & =(2+1-5) \hat{\mathbf{i}}+(1-1+1) \hat{\mathbf{j}}+(-1+0-1) \hat{\mathbf{k}} \\
& =-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}=-(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})
\end{aligned}$
Now, the unit vector in the direction of $\mathbf{a}+\mathbf{b}-\mathbf{c}$ be
$\frac{-(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})}{\sqrt{2^2+(-1)^2+2^2}}=-\frac{1}{3}(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})$
The required unit vector be $\frac{1}{3}(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})$.
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