Let A_1→=3, A_2→=5 and A_1→+A_2→=5. The value of 2A_1→+3A_2→∙3A_1→-2A_2→ is:

Correct answer: (B) -118.5. A→_1+A→_2=5 A→^(1+A→22)=25 A→^12+A→^22+2A→_1A→_2=25 9+25+2A→_1∙A→_2=25 A→_1.A→_2=-92 Now, 2A→_1+3A→_2∙3A→_1-2A→_2=6A→^12-4A→_1A→_2+9A→_1A→_2-6A→^22 =6.A→^12+5A→_1A→_2-6A→^22 =6×9+5-9/2-6×25 =54-45/2-150 = -118.5

Q68 / 90 · JEE Main 2019 (08 Apr Shift 2) · Physics

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:

A
$- 112.5$
Correct answer
B
$- 118.5$
C
$- 106.5$
D
$- 99.5$

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

Hence, the correct answer is option: 'B'.

Let A1→=3, A2→=5 and A1→+A2→=5. The value of 2A1→+3A2→∙3A1→-2A2→ is:

Solution

Relevant Questions

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: