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  • Let <img alt="\overrightarrow{\mathrm{a}}=\alpha \hat{i}+3 \hat{j}-\hat{k}, \overrightarrow{\mathrm{b}}=3 \hat{i}-\beta \hat{j}+4 \hat{k}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cov

Let \overrightarrow{\mathrm{a}}=\alpha \hat{i}+3 \hat{j}-\hat{k}, \overrightarrow{\mathrm{b}}=3 \hat{i}-\beta \hat{j}+4 \hat{k} and \vec{c}=\hat{i}+2 \hat{j}-2 \hat{k}$ where $\alpha, \beta \in \mathbf{R}, be three vectors. If the projection of \vec{a}$ on $\vec{c}$ is $\frac{10}{3}$ and $\vec{b} \times \vec{c}=-6 \hat{i}+10 \hat{j}+7 \hat{k}, then the value of \alpha+\beta is equal to :

Option: 1

3


Option: 2

4


Option: 3

5


Option: 4

6


Answers (1)

best_answer

Projection of   \vec{\mathrm{a}} \text { on } \vec{\mathrm{c}}=\frac{\vec{\mathrm{a}} \cdot \vec{\mathrm{c}}}{|\vec{\mathrm{c}}|}

                                   \mathrm{=\frac{\alpha+6+2}{\sqrt{1+2^{2}+2^{2}}}} \\

                                   \mathrm{=\frac{8+\alpha}{3}}

Given   \mathrm{\frac{8+\alpha}{3}=\frac{10}{3} \Rightarrow \alpha=2}

\vec{\mathrm{b}} \times \vec{\mathrm{c}} =\mathrm{\left|\begin{array}{ccc} i & j & k \\ 3 & -\beta & 4 \\ 1 & 2 & -2 \end{array}\right|} \\

\mathrm{=i(2 \beta-8)-j(-6-4)+k(6+\beta)}

Comparing with given value

\mathrm{2 \beta-8=-6 \Rightarrow 2 \beta=2 \Rightarrow \beta=1} \\

\mathrm{\alpha+\beta=2+1=3}

Hence answer is option 1

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