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If $\vec{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}, \vec{b}=-\hat{i}+2 \hat{j}+\hat{k}$ and $\vec{c}=3 \hat{i}+\hat{j}$ are three
vectors such that $\vec{a}+t \vec{b}$ is perpendicular to $\vec{c}$, then what is t equal to ?
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vectors such that $\vec{a}+t \vec{b}$ is perpendicular to $\vec{c}$, then what is t equal to ?
Solution:
1318 Upvotes
Verified Answer
The correct answer is:
8
$\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=(2-\mathrm{t}) \hat{\mathrm{i}}+(2+2 \mathrm{t}) \hat{\mathrm{j}}+(3+\mathrm{t}) \hat{\mathrm{k}}$
$(\vec{a}+t \vec{b})$ and $\vec{c}$ is perpendicular. Therefore,
$(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \cdot \overrightarrow{\mathrm{c}}=0$
$3(2-\mathrm{t})+2+2 \mathrm{t}=0$
$6-3 \mathrm{t}+2 \mathrm{t}+2=0$
$\mathrm{t}=8$
$(\vec{a}+t \vec{b})$ and $\vec{c}$ is perpendicular. Therefore,
$(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \cdot \overrightarrow{\mathrm{c}}=0$
$3(2-\mathrm{t})+2+2 \mathrm{t}=0$
$6-3 \mathrm{t}+2 \mathrm{t}+2=0$
$\mathrm{t}=8$
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