Let <math> <mrow> <mover accent="true"> <mi>a</mi> <mo>→</mo> </mover> <mtext> </mtext> <mtext> </mtext> <mi>a</mi> <mi>n</mi> <mi>d</mi> <mtext> </mtext> <mtext> </mtext> <mover accent="true"> <mi>b</mi> <mo>→</mo> </mover> </mrow> </math> be two vectors such that <math> <mrow> <mrow> <mo>|</mo> <mrow> <mn>2</mn> <mover accent="true"> <mi>a</mi> <mo>→</mo> </mover> <mo>+</mo> <mn>3</mn> <mover accent="true"> <mi>b</mi> <mo>→</mo> </mover> </mrow> <mo>|</mo> </mrow> <mo>=</mo> <mrow> <mo>|</mo> <mrow> <mn>3</mn> <mover accent="true"> <mi>a</mi> <mo>→</mo> </mover> <mo>+</mo> <mover accent="true"> <mi>b</mi> <mo>→</mo> </mover> </mrow> <mo>|</mo> </mrow> </mrow> </math> and the angle between <math> <mrow> <mover accent="true"> <mi>a</mi> <mo>→</mo> </mover> <mtext> </mtext> <mtext> </mtext> <mi>a</mi> <mi>n</mi> <mi>d</mi> <mtext> </mtext> <mtext> </mtext> <mover accent="true"> <mi>b</mi> <mo>→</mo> </mover> </mrow> </math> is 60°. If <math> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>8</mn> </mrow> </mfrac> <mover accent="true"> <mi>a</mi> <mo>→</mo> </mover> </mrow> </math> is a unit vector, then <math> <mrow> <mrow> <mo>|</mo> <mrow> <mover accent="true"> <mi>b</mi> <mo>→</mo> </mover> </mrow> <mo>|</mo> </mrow> </mrow> </math> is equal to:<math><mrow><mrow><mrow><mover accent="true"></mover></mrow></mrow></mrow></math><math><mrow><mover accent="true"></mover><mover accent="true"></mover></mrow></math>

| 2 a → + 3 b → | 2 = | 3 a → + b → | 2 ⇒ 4 | a → | 2 + 9 | b → | 2 + 1 2 a → . b → = 9 | a → | 2 + | b → | 2 + 6 a → . b → ⇒ − 5 | a → | 2 + 8 | b → | 2 + 6 a → . b → = 0 . . . . . . . . . ( i ) c o s 6 0 ° = a → . b → | a → | | b → | = 1 2 ∴ a → . b → = 4 | b → | a n d | a → | = 8 from (i) we get | b → | = 5

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Maths Class 11th Ncert Solutions Maths class 11th

Let $\vec{a} a n d \vec{b}$ be two vectors such that $| 2 \vec{a} + 3 \vec{b} | = | 3 \vec{a} + \vec{b} |$ and the angle between $\vec{a} a n d \vec{b}$ is 60°. If $\frac{1}{8} \vec{a}$ is a unit vector, then $| \vec{b} |$ is equal to:

Option 1 -

Option 2 -

Option 3 -

Option 4 -

2 Views | Posted 2 months ago

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1 Answer

Answered by

Raj Pandey | Contributor-Level 9

2 months ago

Correct Option - 2

Detailed Solution:

${| 2 \vec{a} + 3 \vec{b} |}^{2} = {| 3 \vec{a} + \vec{b} |}^{2} \Rightarrow 4 {| \vec{a} |}^{2} + 9 {| \vec{b} |}^{2} + 12 \vec{a} . \vec{b} = 9 {| \vec{a} |}^{2} + {| \vec{b} |}^{2} + 6 \vec{a} . \vec{b}$

$\Rightarrow - 5 {| \vec{a} |}^{2} + 8 {| \vec{b} |}^{2} + 6 \vec{a} . \vec{b} = 0 . . . . . . . . . (i)$

$c o s 60 ° = \frac{\vec{a} . \vec{b}}{| \vec{a} | | \vec{b} |} = \frac{1}{2} ∴ \vec{a} . \vec{b} = 4 | \vec{b} | a n d | \vec{a} | = 8$

from (i) we get $| \vec{b} | = 5$

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Let $\vec{a} a n d \vec{b}$ be two vectors such that $| 2 \vec{a} + 3 \vec{b} | = | 3 \vec{a} + \vec{b} |$ and the angle between $\vec{a} a n d \vec{b}$ is 60°. If $\frac{1}{8} \vec{a}$ is a unit vector, then $| \vec{b} |$ is equal to:

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Let a → a n d b → be two vectors such that | 2 a → + 3 b → | = | 3 a → + b → | and the angle between a → a n d b → is 60°. If 1 8 a → is a unit vector, then | b → | is equal to:

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Let $\vec{a} a n d \vec{b}$ be two vectors such that $| 2 \vec{a} + 3 \vec{b} | = | 3 \vec{a} + \vec{b} |$ and the angle between $\vec{a} a n d \vec{b}$ is 60°. If $\frac{1}{8} \vec{a}$ is a unit vector, then $| \vec{b} |$ is equal to: