If a- and a2 are two non collinear unit vectors and if -a-a-3- then the value of a1-a2 -2 a1-a2A- 2 B- 3-2D- 1

Given that a_1 and a_2 are two non collinear unit vectors such that |a_1+a_2|=√(3) |a_1+a_2|^2=(√(3) )^2 (a_1+a_2).(a_1+a_2)=3 nbsp; nbsp;∴ |P|^2=P.P a_1.a ...

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Byju's Answer

Standard XII

Mathematics

Distance Formula

If a⃗ and a2 ...

Question

If $_{1}$ and $_{2}$ are two non collinear unit vectors and if $∣ ∣_{1} +_{2} ∣ ∣ = \sqrt{3}$ , then the value of $(_{1} -_{2}) . (2_{1} +_{2})$ is

2

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\frac{3}{2}

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\frac{1}{2}

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Solution

The correct option is C $\frac{1}{2}$
Given that ${\to a}_{1}$ and ${\to a}_{2}$ are two non-collinear unit vectors such that $| {\to a}_{1} + {\to a}_{2} | = \sqrt{3}$
$| {\to a}_{1} + {\to a}_{2} |^{2} = (\sqrt{3})^{2}$
$({\to a}_{1} + {\to a}_{2}) . ({\to a}_{1} + {\to a}_{2}) = 3$ $∴ | \to P |^{2} = \to P . \to P$
${\to a}_{1} . {\to a}_{1} + {\to a}_{1} . {\to a}_{2} + {\to a}_{2} . {\to a}_{1} + {\to a}_{2} . {\to a}_{2} = 3$
$| {\to a}_{1} |^{2} + 2 ({\to a}_{1} . {\to a}_{2}) + | {\to a}_{2} |^{2} = 3 . . . (i)$

As we know that, $\to P . \to Q = \to Q . \to P$ and also given that $| {\to a}_{1} |, | {\to a}_{2} | = 1$ as it is unit vectors.

Putting the values in equation $(i)$ we get,
$1^{2} + 2 ({\to a}_{1} . {\to a}_{2}) + 1^{2} = 3$ ,
$2 ({\to a}_{1} . {\to a}_{2}) = 1, ({\to a}_{1} . {\to a}_{2}) = \frac{1}{2}$

Hence the value of $(_{1} -_{2}) . (2_{1} +_{2}) = 2 | {\to a}_{1} |^{2} - {\to a}_{1} . {\to a}_{2} - | {\to a}_{2} |^{2}$
$2 (1)^{2} - \frac{1}{2} - (1)^{2} = 2 - \frac{3}{2} = \frac{1}{2}$

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If →a1 and →a2 are two non collinear unit vectors and if ∣∣→a1+→a2∣∣=√3, then the value of (→a1−→a2).(2→a1+→a2) is

If $_{1}$ and $_{2}$ are two non collinear unit vectors and if $∣ ∣_{1} +_{2} ∣ ∣ = \sqrt{3}$ , then the value of $(_{1} -_{2}) . (2_{1} +_{2})$ is