Let A- B- and C- be unit vectors- Suppose that A-B- A-C- 0 and that the angle between B- and C- is -6 then A-

The correct option is A ± 2 ( B⃗×C⃗ ) Since, nbsp; A⃗·B⃗= 0, A⃗·C⃗= 0 Hence, nbsp; ( B⃗+C⃗ )·A⃗= 0 So nbsp; A⃗ is perpendicular to nb ...

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Byju's Answer

Standard XII

Physics

Multiplication with Vectors

Let A⃗, B⃗ ...

Question

Let $\to A, \to B$ and $\to C$ be unit vectors. Suppose that $\to A \cdot \to B = \to A \cdot \to C = 0$ and that the angle between $\to B$ and $\to C$ is $\frac{π}{6}$ then $\to A =$

\pm 2 (\to B \times \to C)

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\pm (\to B \times \to C)

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\pm 2 (\to B + \to C)

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\pm (\to B + \to C)

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Solution

The correct option is A $\pm 2 (\to B \times \to C)$
Since, $\to A \cdot \to B = 0, \to A \cdot \to C = 0$
Hence, $(\to B + \to C) \cdot \to A = 0$
So $\to A$ is perpendicular to $(\to B + \to C) \cdot \to A$ is a unit vector perpendicular to the plane of vectors $\to B$ and $\to C .$
$\to A = \frac{\to B \times \to C}{∣ ∣ \to B \times \to C ∣ ∣}$
$∣ ∣ \to B \times \to C ∣ ∣ = ∣ ∣ \to B ∣ ∣ ∣ ∣ \to C ∣ ∣ sin \frac{π}{6} = 1 \times 1 \times \frac{1}{2} = \frac{1}{2}$
$∴ \to A = \frac{\to B \times \to C}{∣ ∣ \to B \times \to C ∣ ∣} = \pm 2 (\to B \times \to C)$

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Let →A,→B and →C be unit vectors. Suppose that →A⋅→B=→A⋅→C=0 and that the angle between →B and →C is π6 then →A=

Let $\to A, \to B$ and $\to C$ be unit vectors. Suppose that $\to A \cdot \to B = \to A \cdot \to C = 0$ and that the angle between $\to B$ and $\to C$ is $\frac{π}{6}$ then $\to A =$