If a→=2, b→=3 and 2a→-b→=5, then 2a→+b→ equals :

Correct answer: (A) 5. a→=2, b→=3, 2a→-b→=5 2 a→-b→^2=2 a→-b→·2 a→-b→ =4a→^2-4a→·b→+b→^2 ...i 2a→+b→^2=4a→^2+4a→·b→+b→^2 ...ii Adding equation (i) and equation (ii), we get 2 a→+b→^2+2 a→-b→^2=8a→^2+2b→^2 ⇒ 2 a→+b→^2+5^2=82^2+23^2 ⇒ 2a→+b→^2=32+18-25 = 2 5 ⇒ 2a→+b→=5

Q52 / 90 · JEE Main 2014 (09 Apr Online) · Mathematics

If $|\vec{a}| = 2, |\vec{b}| = 3$ and $|2 \vec{a} - \vec{b}| = 5,$ then $|2 \vec{a} + \vec{b}|$ equals :

Correct answer
A
$5$
B
$7$
C
$17$
D
$1$

Solution

$|\vec{a}| = 2, |\vec{b}| = 3, |2 \vec{a} - \vec{b}| = 5$

${|2 \vec{a} - \vec{b}|}^{2} = (2 \vec{a} - \vec{b}) \cdot (2 \vec{a} - \vec{b})$

$= 4 {|\vec{a}|}^{2} - 4 [\vec{a} \cdot \vec{b}] + {|\vec{b}|}^{2} . . . (i)$

${|2 \vec{a} + \vec{b}|}^{2} = 4 {|\vec{a}|}^{2} + 4 [\vec{a} \cdot \vec{b}] + {|\vec{b}|}^{2} . . . (ii)$

Adding equation (i) and equation (ii), we get

${|2 \vec{a} + \vec{b}|}^{2} + {|2 \vec{a} - \vec{b}|}^{2} = 8 {|\vec{a}|}^{2} + {2 |\vec{b}|}^{2}$

$\Rightarrow {|2 \vec{a} + \vec{b}|}^{2} + 5^{2} = 8 (2^{2}) + 2 (3^{2})$

$\Rightarrow {|2 \vec{a} + \vec{b}|}^{2} = 32 + 18 - 25$ $= 25$

$\Rightarrow |2 \vec{a} + \vec{b}| = 5$

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

If a→=2, b→=3 and 2a→-b→=5, then 2a→+b→ equals :

Solution

Relevant Questions

If $|\vec{a}| = 2, |\vec{b}| = 3$ and $|2 \vec{a} - \vec{b}| = 5,$ then $|2 \vec{a} + \vec{b}|$ equals :