Vector c→ is perpendicular to vectors a→=(2,−3,1) and b→=(1,−2,3) and satisfies the condition c→⋅(i^+2j^−7k^)=10 Then vector c→ is equal to

Let c→=λ(a→×b→) Hence, λ(a→×b→)⋅(i^+2j^−7k^)=10⇒ λ2−311−2312−7=10 or λ=−1⇒ c→=−(a→×b→)

Vector c→ is perpendicular to vectors a→=(2,−3,1) and b→=(1,−2,3) and satisfies the condition c→⋅(i^+2j^−7k^)=10 Then vector c→ is equal to

Let c→=λ(a→×b→) Hence, λ(a→×b→)⋅(i^+2j^−7k^)=10⇒ λ2−311−2312−7=10 or λ=−1⇒ c→=−(a→×b→)

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Vector $\vec{c}$ is perpendicular to vectors $\vec{a} = (2, - 3, 1) and \vec{b} = (1, - 2, 3)$ and satisfies the condition $\vec{c} \cdot (\hat{i} + 2 \hat{j} - 7 \hat{k}) = 10$ Then vector $\vec{c}$ is equal to

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(7, 5, 1)

(1, 1, -1)

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(-7,- 5, -1)

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Detailed Solution

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

$Hence, λ (\vec{a} \times \vec{b}) \cdot (\hat{i} + 2 \hat{j} - 7 \hat{k}) = 10$

$\begin{array}{l} \Rightarrow λ |\begin{array}{ccc} 2 & - 3 & 1 \\ 1 & - 2 & 3 \\ 1 & 2 & - 7 \end{array}| = 10 \\ or λ = - 1 \\ \Rightarrow \vec{c} = - (\vec{a} \times \vec{b}) \end{array}$

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Vector c→ is perpendicular to vectors a→=(2,−3,1) and b→=(1,−2,3) and satisfies the condition c→⋅(i^+2j^−7k^)=10 Then vector c→ is equal to

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Vector $\vec{c}$ is perpendicular to vectors $\vec{a} = (2, - 3, 1) and \vec{b} = (1, - 2, 3)$ and satisfies the condition $\vec{c} \cdot (\hat{i} + 2 \hat{j} - 7 \hat{k}) = 10$ Then vector $\vec{c}$ is equal to