Let -A1-3- -A2-5 and -A1-A2-5- The value of 2A1-3A2- 3A1-2A2 is-

Using, |A_1+A_2|=√(A_1^2+A_2^2+2A_1A_2 cosθ) 5=√(3^2+5^2+2 × 3 × 5 cosθ) 25=34+30 cosθ ⇒cosθ= 0.3 Now, (2 A_1+3 A_2) · (3 A_1 2 A_2)= 2 A_1· 3A_1 2 A_1· 2A_ ...

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Byju's Answer

Standard XII

Mathematics

Nature of Roots

Let |A1|=3, |...

Question

Let $|_{1} | = 3, |_{2} | = 5$ and $|_{1} +_{2} | = 5$ . The value of $(2_{1} + 3_{2}) . (3_{1} - 2_{2})$ is,

- 112.5

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- 106.5

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- 99.5

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- 118.5

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Solution

The correct option is D $- 118.5$
Using,
$| {\to A}_{1} + {\to A}_{2} | = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} cos θ}$

$5 = \sqrt{3^{2} + 5^{2} + 2 \times 3 \times 5 cos θ}$

$25 = 34 + 30 cos θ$

$\Rightarrow cos θ = - 0.3$

Now,
$(2_{1} + 3_{2}) \cdot (3_{1} - 2_{2}) = 2_{1} \cdot 3_{1} - 2_{1} \cdot 2_{2} + 3_{1} \cdot 3_{2} - 3_{2} \cdot 2_{2}$

As, $_{1} \cdot_{2} = A_{1} A_{2} cos θ$

$∴ (2_{1} + 3_{2}) \cdot (3_{1} - 2_{2})$

$= 6 A_{1}^{2} - 4 A_{1} A_{2} cos θ + 9 A_{1} A_{2} cos θ - 6 A_{2}^{2}$

$= 6 A_{1}^{2} - 6 A_{2}^{2} + 5 A_{1} A_{2} cos θ$

$= 6 \times 3^{2} - 6 \times 5^{2} + 5 \times 3 \times 5 (- 0.3)$

$= - 118.5$

Hence, $(D)$ is the correct answer.

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Similar questions

Q. Let

| \begin{matrix} A_{1} \end{matrix} | = 3, | \begin{matrix} A_{2} \end{matrix} | = 5,

and

| \begin{matrix} A_{1} + A_{2} \end{matrix} | = 5

A_{1}

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Q. If

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Q. If

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which is not possible is

If $a_{1} + 2 a_{2} + 3 a_{3} = 1$ , where $a_{1}, a_{2} & a_{3}$ are all positive numbers; then what is the minimum value of $2^{a 1} + 4^{a 2} + 8^{a 3}$ ?

Q. Let

a_{1}, a_{2}, . . . . . . . . . a_{11}

be real numbers satisfying

a_{1} = 15, 27 - 2 a_{2} > 0

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a_{k} = 2 a_{k - 1} - a_{k - 2}

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Let |−→A1|=3,|−→A2|=5 and |−→A1+−→A2|=5. The value of (2−→A1+3−→A2).(3−→A1−2−→A2) is,

Let $|_{1} | = 3, |_{2} | = 5$ and $|_{1} +_{2} | = 5$ . The value of $(2_{1} + 3_{2}) . (3_{1} - 2_{2})$ is,