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If $\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}, \vec{b}=-\hat{i}+2 \hat{j}+\hat{k}$ and $\overrightarrow{\mathrm{c}}=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}$ then $t$ such that $\overrightarrow{\mathrm{a}}+\mathrm{t} \overrightarrow{\mathrm{b}}$ is at right angle to $\vec{c}$ will be equal to
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The correct answer is:
5
We have,$\vec{a}+t \vec{b}=(\hat{i}+2 \hat{j}+3 \hat{k})+t(-\hat{i}+2 \hat{j}+\hat{k})$
It is $\perp$ to $\mathrm{c}=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}$
If $3(1-t)+(2+2 t)+(3+t)(0)=0$
$\Rightarrow 3-3 \mathrm{t}+2+2 \mathrm{t}=0 \Rightarrow \mathrm{t}=5$
It is $\perp$ to $\mathrm{c}=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}$
If $3(1-t)+(2+2 t)+(3+t)(0)=0$
$\Rightarrow 3-3 \mathrm{t}+2+2 \mathrm{t}=0 \Rightarrow \mathrm{t}=5$
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