Let a- b and c be three non-coplanar unit vectors such that the angle between each pair of vectors is -3- If a-b-b-c-p a-q b-r c- where p- q and r are scalars- then the value of p2-2 q2-r2q2 is

Given, |a|=|b|=|c|=1 (a×b)+(b×c)=p a+q b+r c Taking dot product with a, we get a·(a×b)+a·(b×c)=p+q(a·b)+r(a·c) ⇒ 0+ [a b c]=p+q2+r2 ⇒ 2p+q+r=2 [a b c] ⋯ ...

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Byju's Answer

Standard XIII

Mathematics

Dot Product of Two Vectors

Let a, b and ...

Question

Let $\to a, \to b$ and $\to c$ be three non-coplanar unit vectors such that the angle between each pair of vectors is $\frac{π}{3} .$ If $\to a \times \to b + \to b \times \to c = p \to a + q \to b + r \to c,$ where $p, q$ and $r$ are scalars, then the value of $\frac{p^{2} + 2 q^{2} + r^{2}}{q^{2}}$ is

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Solution

Given, $| \to a | = | \to b | = | \to c | = 1$
$(\to a \times \to b) + (\to b \times \to c) = p \to a + q \to b + r \to c$
Taking dot product with $\to a,$ we get
$\to a \cdot (\to a \times \to b) + \to a \cdot (\to b \times \to c) = p + q (\to a \cdot \to b) + r (\to a \cdot \to c)$
$\Rightarrow 0 + [\to a \to b \to c] = p + \frac{q}{2} + \frac{r}{2}$ $\Rightarrow 2 p + q + r = 2 [\to a \to b \to c] \dots (1)$

Taking dot product with $\to b,$ we get
$0 = \frac{p}{2} + q + \frac{r}{2}$
$\Rightarrow p + 2 q + r = 0 \dots (2)$

Taking dot product with $\to c,$ we get
$[\to a \to b \to c] + 0 = \frac{p}{2} + \frac{q}{2} + r$ $\Rightarrow p + q + 2 r = 2 [\to a \to b \to c] \dots (3)$

From $(1), (2)$ and $(3),$ we get
$p = - q = r$
Thus, $\frac{p^{2} + 2 q^{2} + r^{2}}{q^{2}} = \frac{p^{2} + 2 p^{2} + p^{2}}{p^{2}} = 4$

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Let →a,→b and →c be three non-coplanar unit vectors such that the angle between each pair of vectors is π3. If →a×→b+→b×→c=p→a+q→b+r→c, where p,q and r are scalars, then the value of p2+2q2+r2q2 is