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  • If <img alt="\vec{a}=2 \hat{i}+\hat{j}+3 \hat{k}, \quad \vec{b}=3 \hat{i}+3 \hat{j}+\hat{k} \: and\; \vec{c}=c_{1} \hat{i}+c_{2} \hat{j}+c_{3} \hat{k}" src="https://entrancecorner.oncodecogs.com/gi

If \vec{a}=2 \hat{i}+\hat{j}+3 \hat{k}, \quad \vec{b}=3 \hat{i}+3 \hat{j}+\hat{k} \: and\; \vec{c}=c_{1} \hat{i}+c_{2} \hat{j}+c_{3} \hat{k} are coplanar
vectors and \vec{a} \cdot \vec{c}=5, \vec{b} \perp \vec{c}, \: then\: 122\left(c_{1}+c_{2}+c_{3}\right) is equal to______.

Option: 1

150


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\vec{a}\cdot \vec{c}= 5\, \Rightarrow 2C_{1}+C_{2}+3C_{3}= 5\, ---(1)
\vec{b}\perp \vec{c}= 3C_{1}+3C_{2}+C_{3}= 0\, ---(2)

\vec{a},\vec{b},\vec{c} \: \text{are coplanar}\mathrm{\Rightarrow \begin{vmatrix} C_{1} & C_{2} &C_{3} \\ 2 & 1 &3 \\ 3& 3 & 1 \end{vmatrix}}= 0
                                        \mathrm{\Rightarrow -8C_{1}+7C_{2}+3C_{3}= 0\, ---(3)}
\mathrm{Eliminating\: \mathrm{C_{3}}\; \; from \: (1) \: \& \: (3)\Rightarrow 10C_{1}-6C_{1}= 5\, ---(4)}
                                   \mathrm{ from \: (2) \: \& \: (3)\Rightarrow 17C_{1}+2C_{2}= 0\, ---(5)}
                                                                  \mathrm{ \Rightarrow 51C_{1}+6C_{2}= 0\, ---(6)}

\mathrm{ \Rightarrow 61C_{1}= 5\; \Rightarrow \; C_{1}= \frac{5}{61},\: C_{2}= \frac{-1}{2}\times 17C_{1}= \frac{-85}{122}.}
\mathrm{ C_{3}= -3\left ( C_{1}+C_{2} \right )}

\mathrm{ So \; C_{1}+C_{2}+C_{3}= -2\left ( C_{1}+C_{2} \right )= \left ( \frac{85}{61}-\frac{10}{61} \right )}

\mathrm{ \Rightarrow 122\left ( C_{1}+C_{2}+C_{3} \right )= 75\times 2= 150}
 

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manish

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