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STD 12 - 10. vector algebra — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 10. vector algebra4 Marks
Question
Let $\vec{a}=2 \hat{i}+5 \hat{j}-\hat{k}, \vec{b}=2 \hat{i}-2 \hat{j}+2 \hat{k}$ and $\overrightarrow{\mathrm{c}}$ be three vectors such that $(\vec{c}+\hat{i}) \times(\vec{a}+\vec{b}+\hat{i})=\vec{a} \times(\vec{c}+\hat{i}) \cdot \vec{a} \cdot \vec{c}=-29$, then $\overrightarrow{\mathrm{c}} \cdot(-2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})$ is equal to :
✓

Answer

b
Let's assume

$\overrightarrow{\mathrm{v}} $$ =\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\hat{\mathrm{i}} $

$ =5 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}}$

and $\overrightarrow{\mathrm{c}}+\hat{\mathrm{i}}=\overrightarrow{\mathrm{p}}$

So,

$ \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{v}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{p}} $

$ \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{v}}+\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{a}}=\overrightarrow{0} $

$ \overrightarrow{\mathrm{p}} \times(\overrightarrow{\mathrm{v}}+\overrightarrow{\mathrm{a}})=\overrightarrow{0} $

$ \Rightarrow \overrightarrow{\mathrm{p}}=\lambda(\overrightarrow{\mathrm{v}}+\overrightarrow{\mathrm{a}}) $

$ \overrightarrow{\mathrm{c}}+\mathrm{i}=\lambda(7 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}) $

$ \overline{\mathrm{a}} \cdot \overline{\mathrm{c}}+\overline{\mathrm{a}} \cdot \hat{\mathrm{i}}=\lambda \overline{\mathrm{a}} \cdot(7 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}) $

$ -29+2=\lambda(14+40) $

$ \lambda=-\frac{1}{2}$

$ \overrightarrow{\mathrm{c}} \cdot(-2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})+\hat{\mathrm{i}} \cdot(-2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})=\lambda(7 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}) \cdot(-2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}) $

$ \quad=-\frac{1}{2}(-14+8)+2=5$

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