MHT CET · Maths · Vector Algebra
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
- A 10, 1, 4
- B \(10,-4,1\)
- C 10, \(-1,-4\)
- D \(10,1,-4\)
Answer & Solution
Correct Answer
(C) 10, \(-1,-4\)
Step-by-step Solution
Detailed explanation
\(
\begin{array}{l}
\vec{a}=\hat{\imath}+\hat{\jmath}+\hat{k} \\
\vec{b}=2 \hat{\imath}-2 \hat{\jmath}+2 \hat{k} \\
\vec{c}=2 \hat{\imath}+3 \hat{\jmath}+2 \hat{k} \\
x \vec{a}+y \vec{b}=\vec{c}
\end{array}
\)
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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