The vectors a and b are not perpendicular and c and d are two vectors satisfyingb-c - b-d and a - d - 0- Then- the vector d is equal to-

Explanation for correct option The vectors a and b are not perpendicularSo, a.b 0Given, a.d = 0b × c = b × d 0ex0ex∴a× (b×c) = a× (b×d)0ex0ex⇒ (a·c)b – (a·b ...

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Byju's Answer

Standard XII

Mathematics

Cross Product of Two Vectors

The vectors a...

Question

The vectors $\vec{a}$ and $\vec{b}$ are not perpendicular and $\vec{c}$ and $\vec{d}$ are two vectors satisfying $\vec{b} \times \vec{c} = \vec{b} \times \vec{d}$ and $\vec{a} \cdot \vec{d} = 0$ . Then, the vector $\vec{d}$ is equal to?

$\vec{c} + (\frac{\vec{a} \cdot \vec{c}}{\vec{a} \cdot \vec{d}}) \vec{b}$

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$\vec{b} + (\frac{\vec{b} \cdot \vec{c}}{\vec{a} \cdot \vec{b}}) \vec{c}$

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

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

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Solution

The correct option is C
$\vec{c} - (\frac{\vec{a} \cdot \vec{c}}{\vec{a} \cdot \vec{b}}) \vec{b}$

Explanation for correct option
The vectors $\vec{a}$ and $\vec{b}$ are not perpendicular
So, $\vec{a} . \vec{b} \neq 0$
Given, $\vec{a} . \vec{d} = 0$
$b \times c = b \times d ∴ \vec{a} \times (\vec{b} \times \vec{c}) = \vec{a} \times (\vec{b} \times \vec{d}) \Rightarrow (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} = (\vec{a} \cdot d) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{d} \Rightarrow (\vec{a} \cdot \vec{b}) \vec{d} = - (\vec{a} . \vec{c}) \vec{b} + (\vec{a} . \vec{b}) \vec{c} \Rightarrow \vec{d} = \vec{c} - (\frac{\vec{a} . \vec{c}}{\vec{a} \cdot \vec{b}}) \vec{b}$
Hence, option (C) is correct

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The vectors a→ and b→ are not perpendicular and c→ and d→ are two vectors satisfyingb→×c→=b→×d→ and a→·d→=0. Then, the vector d→ is equal to?

The vectors $\vec{a}$ and $\vec{b}$ are not perpendicular and $\vec{c}$ and $\vec{d}$ are two vectors satisfying $\vec{b} \times \vec{c} = \vec{b} \times \vec{d}$ and $\vec{a} \cdot \vec{d} = 0$ . Then, the vector $\vec{d}$ is equal to?