SATHEE: Vectors 2 Question 20

Vectors 2 Question 20

20. Let $\vec{A}, \vec{B}$ and $\vec{C}$ be unit vectors. If $\vec{A} \cdot \vec{B} = \vec{A} \cdot \vec{C} = 0$ and that the angle between $\vec{B}$ and $\vec{C}$ is $π / 6$ .

Then, $\vec{A} = \pm 2 (\vec{B} \times \vec{C})$ .

(1981, 2M)

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Solution:

Given, $\vec{A} \cdot \vec{B} = \vec{A} \cdot \vec{C} = 0$

$\Rightarrow \vec{A}$ is perpendicular to both $\vec{B}$ and $\vec{C}$ .

$\Rightarrow \vec{A} = λ (\vec{B} \times \vec{C})$

$| \vec{A} | = | λ | | \vec{B} \times \vec{C} |$ , where $\vec{A}, \vec{B}, \vec{C}$ are unit vectors.

$\Rightarrow | λ | = \frac{1}{1 \cdot \sin 30^{\circ}} \Rightarrow | λ | = 2 \Rightarrow λ = \pm 2$

$∴ \vec{A} = \pm 2 (\vec{B} \times \vec{C})$

Hence, given statement is true.

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Vectors 2 Question 20

20. Let A→,B→ and C→ be unit vectors. If A→⋅B→=A→⋅C→=0 and that the angle between B→ and C→ is π/6.

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20. Let $\vec{A}, \vec{B}$ and $\vec{C}$ be unit vectors. If $\vec{A} \cdot \vec{B} = \vec{A} \cdot \vec{C} = 0$ and that the angle between $\vec{B}$ and $\vec{C}$ is $π / 6$ .