SATHEE: Vectors 4 Question 16

Vectors 4 Question 16

16. If $\vec{a}$ and $\vec{b}$ are vectors in space given by $\vec{a} = \frac{\hat{i} - 2 \hat{j}}{\sqrt{5}}$ and $\vec{b} = \frac{2 \hat{i} + \hat{j} + 3 \hat{k}}{\sqrt{14}}$ , then the value of

$(2 \vec{a} + \vec{b}) \cdot [(\vec{a} \times \vec{b}) \times (\vec{a} - 2 \vec{b})]$ is…

(2010)

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Solution:

From the given information, it is clear that

$\vec{a} = \frac{\hat{i} - 2 \hat{j}}{\sqrt{5}} \Rightarrow | \vec{a} | = 1, | \vec{b} | = 1, \vec{a} \cdot \vec{b} = 0$

Now, $(2 \vec{a} + \vec{b}) \cdot [(\vec{a} \times \vec{b}) \times (\vec{a} - 2 \vec{b})]$

$\begin{aligned} = (2 \vec{a} + \vec{b}) \cdot [a^{2} \vec{b} - (\vec{a} \cdot \vec{b}) \cdot \vec{a} + 2 b^{2} \cdot \vec{a} - 2 (\vec{b} \cdot \vec{a}) \cdot b] \\ = [2 \vec{a} + \vec{b}] \cdot [\vec{b} + 2 \vec{a}] = 4 {\vec{a}}^{2} + {\vec{b}}^{2} \\ = 4 \cdot 1 + 1 = 5 [as a \cdot b = 0] \end{aligned}$

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Vectors 4 Question 16

16. If a→ and b→ are vectors in space given by a→=i^−2j^5 and b→=2i^+j^+3k^14, then the value of

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16. If $\vec{a}$ and $\vec{b}$ are vectors in space given by $\vec{a} = \frac{\hat{i} - 2 \hat{j}}{\sqrt{5}}$ and $\vec{b} = \frac{2 \hat{i} + \hat{j} + 3 \hat{k}}{\sqrt{14}}$ , then the value of