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If $\bar{a}=\hat{\imath}+\hat{j}+\hat{k}, \bar{b}=2 \hat{\imath}-2 \hat{\jmath}+2 \hat{k}, \bar{c}=2 \hat{\imath}+3 \hat{\jmath}+2 \hat{k}$ are any three co-planar vectors such that $l \bar{a}+m \bar{b}+n \bar{c}=\overline{0}$, then values of $l, m, n$ are respectively
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1134 Upvotes
Verified Answer
The correct answer is:
10, $-1,-4$
$$
\begin{array}{l}
\vec{a}=\hat{\imath}+\hat{\jmath}+\hat{k} \\
\vec{b}=2 \hat{\imath}-2 \hat{\jmath}+2 \hat{k} \\
\vec{c}=2 \hat{\imath}+3 \hat{\jmath}+2 \hat{k} \\
x \vec{a}+y \vec{b}=\vec{c}
\end{array}
$$
By solving all eq"s we get $x=+\frac{5}{2} ; y=-\frac{1}{4}$
$$
10 \vec{a}-\vec{b}-4 \vec{c}=0
$$
\begin{array}{l}
\vec{a}=\hat{\imath}+\hat{\jmath}+\hat{k} \\
\vec{b}=2 \hat{\imath}-2 \hat{\jmath}+2 \hat{k} \\
\vec{c}=2 \hat{\imath}+3 \hat{\jmath}+2 \hat{k} \\
x \vec{a}+y \vec{b}=\vec{c}
\end{array}
$$
By solving all eq"s we get $x=+\frac{5}{2} ; y=-\frac{1}{4}$
$$
10 \vec{a}-\vec{b}-4 \vec{c}=0
$$
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