Let A 1 → = 3 , A 2 → = 5 and A 1 → + A 2 → = 5 . The value of 2 A 1 → + 3 A 2 → ∙ 3 A 1 → - 2 A 2 → is:

A → 1 + A → 2 = 5 A → 1 + A → 2 2 = 25 A → 1 2 + A → 2 2 + 2 A → 1 A → 2 = 25 9 + 25 + 2 A → 1 ∙ A → 2 = 25 A → 1 . A → 2 = - 9 2 Now, 2 A → 1 + 3 A → 2 ∙ 3 A → 1 - 2 A → 2 = 6 A → 1 2 - 4 A → 1 A → 2 + 9 A → 1 A → 2 - 6 A → 2 2 = 6 . A → 1 2 + 5 A → 1 A → 2 - 6 A → 2 2 = 6 × 9 + 5 - 9 2 - 6 × 25 = 54 - 45 2 - 150 = - 118.5

Physics - Vectors Question with Solution | TestHub

PhysicsVectorsMultiplication of vectorsEasy2 minPYQ_2019

PhysicsEasysingle choice

Let $|\vec{A_{1}}| = 3, |\vec{A_{2}}| = 5$ and $|\vec{A_{1}} + \vec{A_{2}}| = 5 .$ The value of $(2 \vec{A_{1}} + 3 \vec{A_{2}}) ∙ (3 \vec{A_{1}} - 2 \vec{A_{2}})$ is:

Options:

Answer:

Solution:

$|{\vec{A}}_{1} + {\vec{A}}_{2}| = 5$
${|{\vec{A}}_{1} + {\vec{A}}_{2}|}^{2} = 25$
${|{\vec{A}}_{1}|}^{2} + {|{\vec{A}}_{2}|}^{2} + 2 {\vec{A}}_{1} {\vec{A}}_{2} = 25$
$9 + 25 + 2 {\vec{A}}_{1} ∙ {\vec{A}}_{2} = 25$
${\vec{A}}_{1} . {\vec{A}}_{2} = - \frac{9}{2}$
Now, $(2 {\vec{A}}_{1} + 3 {\vec{A}}_{2}) ∙ (3 {\vec{A}}_{1} - 2 {\vec{A}}_{2}) = 6 {|{\vec{A}}_{1}|}^{2} - 4 {\vec{A}}_{1} {\vec{A}}_{2} + 9 {\vec{A}}_{1} {\vec{A}}_{2} - 6 {|{\vec{A}}_{2}|}^{2}$
$= 6 . {|{\vec{A}}_{1}|}^{2} + 5 {\vec{A}}_{1} {\vec{A}}_{2} - 6 {|{\vec{A}}_{2}|}^{2}$
$= 6 \times 9 + 5 (- \frac{9}{2}) - 6 \times 25$
$= 54 - \frac{45}{2} - 150$
$= - 118.5$

Stream:JEESubject:PhysicsTopic:VectorsSubtopic:Multiplication of vectors

⏱ 2mℹ️ Source: PYQ_2019

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