Let , and be three vectors such that is coplanar with and . If the vector is perpendicular to and , then is equal to [2025]

Question

Let $\vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ , $\vec{b} = 3 \hat{i} + \hat{j} - \hat{k}$ and $\vec{c}$ be three vectors such that $\vec{c}$ is coplanar with $\vec{a}$ and $\vec{b}$ . If the vector $\vec{c}$ is perpendicular to $\vec{b}$ and $\vec{a} \cdot \vec{c} = 5$ , then $| \vec{c} |$ is equal to [2025]

Smart Achievers · Accepted Answer

(1)

$| \vec{b} | = \sqrt{9 + 1 + 1} = \sqrt{11}, \vec{a} \cdot \vec{b} = 3 + 2 - 3 = 2$

As $\vec{c}$ is perpendicular to $\vec{b}$ and coplanar with $\vec{a}$ and $\vec{b}$ .

$∴ \vec{c} = λ (\vec{b} \times (\vec{a} \times \vec{b}))$

$= λ ((\vec{b} \cdot \vec{b}) \vec{a} - (\vec{a} \cdot \vec{b}) \vec{b}) = λ (11 \vec{a} - 2 \vec{b})$

$\Rightarrow \vec{c} = λ (11 \hat{i} + 22 \hat{j} + 33 \hat{k} - 6 \hat{i} - 2 \hat{j} + 2 \hat{k}) = 5 λ (\hat{i} + 4 \hat{j} + 7 \hat{k})$

As $\vec{a} \cdot \vec{c} = 5 \Rightarrow 5 λ (1 + 8 + 21) = 5$

$\Rightarrow λ = \frac{1}{30} ∴ \vec{c} = \frac{1}{6} (\hat{i} + 4 \hat{j} + 7 \hat{k})$

$\Rightarrow$ $| \vec{c} |$ $= \frac{1}{6} \sqrt{1 + 16 + 49} = \sqrt{\frac{11}{6}}$

Solution

1 $\sqrt{\frac{11}{6}}$

2 16

3 $\frac{1}{3 \sqrt{2}}$

4 18

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Solution

Q. Let a→=i^+2j^+3k^, b→=3i^+j^–k^ and c→ be three vectors such that c→ is coplanar with a→ and b→. If the vector c→ is perpendicular to b→ and a→·c→=5, then |c→| is equal to [2025]

1 116

2 16

3 132

4 18

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1 $\sqrt{\frac{11}{6}}$

3 $\frac{1}{3 \sqrt{2}}$