Let a ? = i ^ + 2 j ^ + k ^ , b ? = 3 ( i ^ – j ^ + k ^ ) . Let c ? be the vector such that a ? × c ? = b ? and a ? · c ? = 3 . Then a ? · ( ( c ? × b ? ) – b ? – c ? ) is equal to : [2024]

Question

Let $\vec{a} = \hat{i} + 2 \hat{j} + \hat{k}$ , $\vec{b} = 3 (\hat{i} - \hat{j} + \hat{k})$ . Let $\vec{c}$ be the vector such that $\vec{a} \times \vec{c} = \vec{b}$ and $\vec{a} \cdot \vec{c} = 3$ . Then $\vec{a} \cdot ((\vec{c} \times \vec{b}) - \vec{b} - \vec{c})$ is equal to : [2024]

Smart Achievers · Accepted Answer

(3)

$\vec{a} \cdot ((\vec{c} \times \vec{b}) - \vec{b} - \vec{c}) = \vec{a} \cdot (\vec{c} \times \vec{b}) - \vec{a} \cdot \vec{b} - \vec{a} \cdot \vec{c}$

[Using distributive property of multiplication]

$= [\vec{a} \vec{c} \vec{b}] - \vec{a} \cdot \vec{b} - \vec{a} \cdot \vec{c}$ ... (i)

Now, $\vec{a} \times \vec{c} = \vec{b} \Rightarrow (\vec{a} \times \vec{c}) \cdot \vec{b} = \vec{b} \cdot \vec{b}$

$\Rightarrow [\vec{a} \vec{c} \vec{b}] = | \vec{b} |^{2} = {(\sqrt{{(3)}^{2} + {(- 3)}^{2} + {(3)}^{2}})}^{2} = 27$

$\Rightarrow [\vec{a} \vec{c} \vec{b}] = 27$ ... (ii)

$\vec{a} \cdot \vec{b} = 3 (\hat{i} + 2 \hat{j} + \hat{k}) (\hat{i} - \hat{j} + \hat{k}) = 3 (1 - 2 + 1) = 0$ (iii)

and $\vec{a} \cdot \vec{c} = 3$ ... (iv)

Using (ii), (iii) and (iv) in (i), we get

$\vec{a} \cdot ((\vec{c} \times \vec{b}) - \vec{b} - \vec{c}) = 27 - 0 - 3 = 24$

Solution

Q.
Let $\vec{a} = \hat{i} + 2 \hat{j} + \hat{k}$ , $\vec{b} = 3 (\hat{i} - \hat{j} + \hat{k})$ . Let $\vec{c}$ be the vector such that $\vec{a} \times \vec{c} = \vec{b}$ and $\vec{a} \cdot \vec{c} = 3$ . Then $\vec{a} \cdot ((\vec{c} \times \vec{b}) - \vec{b} - \vec{c})$ is equal to : [2024]

1 36

2 20

3 24

4 32

Want Us to Email you About Special Offers & Updates?