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Let $\vec{a}=(1,-2,3)$ and $\vec{b}=(3,1,2)$ be two vectors and $\vec{c}$ be a
vector of length $l$ and parallel to $(\vec{a}+\vec{b})$. What is $\vec{c}$ equal
to?
Options:
vector of length $l$ and parallel to $(\vec{a}+\vec{b})$. What is $\vec{c}$ equal
to?
Solution:
2103 Upvotes
Verified Answer
The correct answer is:
None of these
Given $\overrightarrow{\mathbf{a}}=\overrightarrow{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$
and $\overrightarrow{\mathbf{b}}=3 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$
$\therefore \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}=4 \hat{\mathbf{i}}-\mathbf{j}+5 \hat{\mathbf{k}}$
Then, $\overrightarrow{\mathbf{c}}=\lambda(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}})$
$=\lambda(4 \hat{\mathbf{i}}-\hat{\mathbf{j}}+5 \hat{\mathbf{k}})$
$\Rightarrow l=\sqrt{16 \lambda^{2}+\lambda^{2}+25 \lambda^{2}}$
$\Rightarrow l=\sqrt{42} \lambda$
$\lambda=\frac{l}{\sqrt{42}}$
$\therefore \quad \overrightarrow{\mathbf{c}}=\frac{l}{\sqrt{42}}(4 \hat{\mathbf{i}}-\hat{\mathbf{j}}+5 \hat{\mathbf{k}})=\frac{1}{\sqrt{42}}(4,-1,5)$
and $\overrightarrow{\mathbf{b}}=3 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$
$\therefore \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}=4 \hat{\mathbf{i}}-\mathbf{j}+5 \hat{\mathbf{k}}$
Then, $\overrightarrow{\mathbf{c}}=\lambda(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}})$
$=\lambda(4 \hat{\mathbf{i}}-\hat{\mathbf{j}}+5 \hat{\mathbf{k}})$
$\Rightarrow l=\sqrt{16 \lambda^{2}+\lambda^{2}+25 \lambda^{2}}$
$\Rightarrow l=\sqrt{42} \lambda$
$\lambda=\frac{l}{\sqrt{42}}$
$\therefore \quad \overrightarrow{\mathbf{c}}=\frac{l}{\sqrt{42}}(4 \hat{\mathbf{i}}-\hat{\mathbf{j}}+5 \hat{\mathbf{k}})=\frac{1}{\sqrt{42}}(4,-1,5)$
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