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  • Let <img alt="\mathrm{\vec{a}=\hat{i}+\hat{j}+2 \hat{k}, \vec{b}=2 \hat{i}-3 \hat{j}+\hat{k}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cvec%7Ba%7D%3D%5Chat%7Bi%7D&plu

Let \mathrm{\vec{a}=\hat{i}+\hat{j}+2 \hat{k}, \vec{b}=2 \hat{i}-3 \hat{j}+\hat{k}} and \mathrm{\vec{c}=\hat{i}-\hat{j}+\hat{k}} be three given vectors. Let \mathrm{\vec{v} } be a vector in the plane of \mathrm{\vec{a} } and \mathrm{\vec{b} } whose projection on \mathrm{\vec{c}\; \text{is}\; \frac{2}{\sqrt{3}} }. If \mathrm{\vec{v} \cdot \hat{j}=7 }, then \mathrm{\vec{v} \cdot(\hat{i}+\hat{k})} is equal to :

Option: 1

6


Option: 2

7


Option: 3

8


Option: 4

9


Answers (1)

best_answer

\begin{aligned} &\bar{v}=\lambda \bar{a}+\mu \bar{b}\\ &\bar{v}=(\lambda+2 k) \hat{\imath}+(\lambda-3 \mu) \hat{\jmath}+(2 \lambda+k) \hat{k}\\ &\text { v. } \frac{\vec{c}}{|\vec{c}|}=\frac{2}{\sqrt{3}}\\ &\begin{array}{r} \operatorname{als} 0 \overline{5}, j=7 \\ \geq \lambda-2 \mu=7 \end{array}\\ &\lambda+2 \mu-\lambda+3 \mu+2 \lambda+\mu=2\\ &2 \lambda+6 \mu=2\\ &x+3 \mu=1\\ &\therefore \lambda=4, \mu=-1\\ &\therefore \bar{j} \cdot(\hat{i}+\hat{k})=(2 \vec{i}+\vec{j}+7 \hat{k}) \cdot(\hat{i}+\hat{k})\\ &=9 \end{aligned}

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Pankaj Sanodiya

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