Let a = - i ^ - j ^ + k ^ , a · b = 1 and a × b = i ^ - j ^ . Then a - 6 b is equal to [2023]

Question

Let $\vec{a} = - \hat{i} - \hat{j} + \hat{k},$ $\vec{a} \cdot \vec{b} = 1$ and $\vec{a} \times \vec{b} = \hat{i} - \hat{j} .$ Then $\vec{a} - 6 \vec{b}$ is equal to [2023]

Smart Achievers · Accepted Answer

(4)

$\vec{a} = - \hat{i} - \hat{j} + \hat{k},$ $\vec{a} \cdot \vec{b} = 1,$ $\vec{a} \times \vec{b} = \hat{i} - \hat{j}$

$Let \vec{b} = x \hat{i} + y \hat{j} + z \hat{k}$

$Now, \vec{a} \cdot \vec{b} = 1$

$\Rightarrow - x - y + z = 1$ ...(i)

also, $\vec{a} \times \vec{b} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ - 1 & - 1 & 1 \\ x & y & z \end{matrix} |$ $= \hat{i} (- z - y) - \hat{j} (- z - x) + \hat{k} (- y + x)$

$\Rightarrow \hat{i} (- z - y) - \hat{j} (- z - x) + \hat{k} (- y + x) = \hat{i} - \hat{j} (∵ \vec{a} \times \vec{b} = \hat{i} - \hat{j})$

$- z - y = 1, z + x = - 1, - y + x = 0$ ...(ii)

$Solving (i) & (ii) for x, y, z$

We get $x = \frac{- 2}{3}, y = \frac{- 2}{3}, z = \frac{- 1}{3}$

$So, \vec{a} - 6 \vec{b} = - \hat{i} - \hat{j} + \hat{k} - 6 (\frac{- 2}{3} \hat{i} - \frac{2}{3} \hat{j} - \frac{1}{3} \hat{k})$

$\vec{a} - 6 \vec{b} = 3 (\hat{i} + \hat{j} + \hat{k})$

Solution

Q.
Let $\vec{a} = - \hat{i} - \hat{j} + \hat{k},$ $\vec{a} \cdot \vec{b} = 1$ and $\vec{a} \times \vec{b} = \hat{i} - \hat{j} .$ Then $\vec{a} - 6 \vec{b}$ is equal to [2023]

1 $3 (\hat{i} - \hat{j} - \hat{k})$

2 $3 (\hat{i} - \hat{j} + \hat{k})$

3 $3 (\hat{i} + \hat{j} - \hat{k})$

4 $3 (\hat{i} + \hat{j} + \hat{k})$

Want Us to Email you About Special Offers & Updates?

Solution

Q. Let a→=-i^-j^+k^, a→·b→=1 and a→×b→=i^-j^. Then a→-6b→ is equal to [2023]

1 3(i^-j^-k^)

2 3(i^-j^+k^)

3 3(i^+j^-k^)