If a- b- and c are unit vectors such that a-2b -2c -0- then -a-c- is equal to -

A+2b +2c = 0 ⇒ nbsp;a+2c = 2b ⇒ (a+2c)· (a+2c) = ( 2b )· ( 2b ) ⇒ 1 + 4 nbsp;a· nbsp;c +4= 4 ⇒ nbsp;a· nbsp;c = 14 Let the angle between a and c be ...

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Byju's Answer

Standard XI

Mathematics

Cross Product of Two Vectors

If a, b, and ...

Question

If $\to a, \to b,$ and $\to c$ are unit vectors such that $\to a + 2 \to b + 2 \to c = \to 0,$ then $| \to a \times \to c |$ is equal to :

\frac{\sqrt{15}}{4}

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\frac{1}{4}

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\frac{15}{16}

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\frac{\sqrt{15}}{16}

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Solution

The correct option is A $\frac{\sqrt{15}}{4}$
$\to a + 2 \to b + 2 \to c = \to 0 \Rightarrow \to a + 2 \to c = - 2 \to b \Rightarrow (\to a + 2 \to c) \cdot (\to a + 2 \to c) = (- 2 \to b) \cdot (- 2 \to b) \Rightarrow 1 + 4 \to a \cdot \to c + 4 = 4 \Rightarrow \to a \cdot \to c = - \frac{1}{4}$
Let the angle between $\to a and \to c be θ$
$\Rightarrow cos θ = - \frac{1}{4} \Rightarrow sin θ = \pm \frac{\sqrt{15}}{4}$
Now,
$| \to a \times \to c | = | \to a | | \to c | sin θ = \frac{\sqrt{15}}{4}$

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If →a,→b, and →c are unit vectors such that →a+2→b+2→c=→0, then |→a×→c| is equal to :

The correct option is A √154→a+2→b+2→c=→0⇒→a+2→c=−2→b⇒(→a+2→c)⋅(→a+2→c)=(−2→b)⋅(−2→b)⇒1+4→a⋅→c+4=4⇒→a⋅→c=−14 Let the angle between →a and →c be θ ⇒cosθ=−14⇒sinθ=±√154 Now, |→a×→c|=|→a||→c|sinθ=√154

If $\to a, \to b,$ and $\to c$ are unit vectors such that $\to a + 2 \to b + 2 \to c = \to 0,$ then $| \to a \times \to c |$ is equal to :