Let a ? = i ^ + j ^ + k ^ , b ? = 2 i ^ + 4 j ^ – 5 k ^ and c ? = x i ^ + 2 j ^ + 3 k ^ , x ? R . If d ? is the unit vector in the direction of b ? + c ? such that a ? · d ? = 1 , then ( a ? × b ? ) · c ? is equal to [2024]

Question

Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ , $\vec{b} = 2 \hat{i} + 4 \hat{j} - 5 \hat{k}$ and $\vec{c} = x \hat{i} + 2 \hat{j} + 3 \hat{k}, x \in R$ . If $\vec{d}$ is the unit vector in the direction of $\vec{b} + \vec{c}$ such that $\vec{a} \cdot \vec{d} = 1$ , then $(\vec{a} \times \vec{b}) \cdot \vec{c}$ is equal to [2024]

Smart Achievers · Accepted Answer

(1)

$\vec{a} = \hat{i} + \hat{j} + \hat{k}$ , $\vec{b} = 2 \hat{i} + 4 \hat{j} - 5 \hat{k}$ and $\vec{c} = x \hat{i} + 2 \hat{j} + 3 \hat{k}, x \in R$

$\vec{b} + \vec{c} = (x + 2) \hat{i} + 6 \hat{j} - 2 \hat{k} a n d$ $| (\vec{b} + \vec{c}) |$

$= \sqrt{{(x + 2)}^{2} + {(6)}^{2} + {(- 2)}^{2}} = \sqrt{{(x + 2)}^{2} + 40}$

Now, $\vec{d} = \frac{x + 2}{\sqrt{40 + {(x + 2)}^{2}}} \hat{i} + \frac{6}{\sqrt{40 + {(x + 2)}^{2}}} \hat{j} - \frac{2}{\sqrt{40 + {(x + 2)}^{2}}} \hat{k}$

Given that $\vec{a} \cdot \vec{d} = 1$

$\Rightarrow \frac{x + 2 + 6 - 2}{\sqrt{40 + {(x + 2)}^{2}}} = 1$

$\Rightarrow x + 6 = \sqrt{40 + {(x + 2)}^{2}}$

$\Rightarrow {(x + 6)}^{2} = 40 + {(x + 2)}^{2}$

$\Rightarrow x^{2} + 36 + 12 x = 40 + x^{2} + 4 + 4 x$

$\Rightarrow x = 1$

Now, $(\vec{a} \times \vec{b})$ $\cdot \vec{c} =$ $[\vec{a} \vec{b} \vec{c}]$ $=$ $| \begin{matrix} 1 & 1 & 1 \\ 2 & 4 & - 5 \\ 1 & 2 & 3 \end{matrix} |$

= 1(12 + 10) – 1(6 + 5) + 1(4 – 4) = 11.

Solution

1 11

2 3

3 6

4 9

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Solution

Q. Let a→=i^+j^+k^, b→=2i^+4j^–5k^ and c→=xi^+2j^+3k^, x∈R. If d→ is the unit vector in the direction of b→+c→ such that a→·d→=1, then (a→×b→)·c→ is equal to [2024]

1 11

2 3

3 6

4 9

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