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  • Let <img alt="\vec{a}=4 \hat{i}+3 \hat{j}+5 \hat{k}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cvec%7Ba%7D%3D4%20%5Chat%7Bi%7D&plus;3%20%5Chat%7Bj%7D&plus;5%20%5Chat%7Bk%7

Let \vec{a}=4 \hat{i}+3 \hat{j}+5 \hat{k} and \vec{\beta}=\hat{i}+2 \hat{j}-4 \hat{k}, Let \vec{\beta}_1 be parallel to \vec{a} and \vec{\beta}_2 be perpendicular to \vec{a}.If \vec{\beta}=\vec{\beta}_{1}+\vec{\beta}_{2}, then the value of  5 \vec{\beta}_2 \cdot(\hat{i}+\hat{j}+\hat{k}) is

Option: 1

7


Option: 2

9


Option: 3

6


Option: 4

11


Answers (1)

best_answer

\overrightarrow{\beta_1}=\frac{(\vec{\alpha} \cdot \vec{\beta})}{|\vec{\alpha}|} \hat{\alpha}

\begin{aligned} & =\left(\frac{4+6-20}{\sqrt{16+9+25}}\right) \frac{(4,3,5)}{\sqrt{50}} \\ & =\frac{-10}{50}(4,3,5) \end{aligned}

\begin{aligned} & \vec{\beta}_1=\frac{(-4,-3,-5)}{5} \\ & \vec{\beta}_1+\vec{\beta}_2-=(1,2,-4) \\ & \beta_2=\left(1+\frac{4}{5}, 2+\frac{3}{5},-4+1\right) \end{aligned}

\begin{aligned} & \beta_2=\left(\frac{9}{5}, \frac{13}{5},-3\right) \\ & \therefore 5 \beta_2=(9,13,-15) \\ & \therefore 5 \beta_2 \cdot(1,1,1)=9+13-15 \\ & =7 \end{aligned}

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