Ask & Answer Question

If $\vec{a}, \vec{b}, \vec{c}$ are non-zero and $\vec{b}$ and $\vec{c}$ are non-collinear. $\vec{a} + 5 \vec{b}$ is collinear with $\vec{c}$ and $\vec{b} + 6 \vec{c}$ is collinear with $\vec{a}$ . If $\vec{a} + α \vec{b} + β \vec{b} = 0$ , then find α + β.

3 Views | Posted a month ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

a month ago

$\vec{a} + 5 \vec{b}$ is collinear with $\vec{c}$

⇒ $\vec{a} + 5 \vec{b}$ = $\vec{c}$ …(1)

$\vec{b} + 6 \vec{c}$ is collinear with $\vec{a}$

⇒ $\vec{b} + 6 \vec{c} = μ \vec{a}$ …(2)

From (1) and (2)

$\vec{b} + 6 \vec{c} = μ (λ \vec{c} - 5 \vec{b})$

-> $(1 + 5 μ) \vec{b} + (6 - λ μ) \vec{c} = 0$

$? \vec{b}$ and $\vec{c}$ are non-collinear

-> 1 + 5m = 0 $μ = \frac{- 1}{5}$ and 6 – lm = 0 Þ lm = 6

-> l = – 30

Now,

$\vec{b} = 6 \vec{c} = \frac{- 1}{5} \vec{a}$

$5 \vec{b} + 30 \vec{c} = - \vec{a}$

$\begin{matrix} \vec{a} + 5 \vec{b} + 30 \vec{c} = 0 & \vec{a} + α \vec{b} + β \vec{c} = 0 \end{matrix}]$

On comparing

α = 5, β = 30 ⇒ α + β = 35

<math> <mrow> <mover accent="true"> <mi>a</mi> <mo>→</mo> </mover> <mo>+</mo> <mn>5</mn> <mover accent="true"> <mi>b</mi> <mo>→</mo> </mover> </mrow> </math>  is collinear with <math> <mrow> <mover accent="true"> <mi>c</mi> <mo>→</mo> </mover> </mrow> </math>  ⇒   <math> <mrow> <mover accent="true"> <mi>a</mi> <mo>→</mo> </mover> <mo>+</mo> <mn>5</mn> <mover accent="true"> <mi>b</mi> <mo>→</mo> </mover> </mrow> </math> = <math> <mrow> <mover accent="true"> <mi>c</mi> <mo>→</mo> </mover> </mrow> </math>           …(1) <math> <mrow> <mover accent="true"> <mi>b</mi> <mo>→</mo> </mover> <mo>+</mo> <mn>6</mn> <mover accent="true"> <mi>c</mi> <mo>→</mo> </mover> </mrow> </math> is collinear with <math> <mrow> <mover accent="true"> <mi>a</mi> <mo>→</mo> </mover> </mrow> </math>  ⇒   <math> <mrow> <mover accent="true"> <mi>b</mi> <mo>→</mo> </mover> <mo>+</mo> <mn>6</mn> <mover accent="true"> <mi>c</mi> <mo>→</mo> </mover> <mo>=</mo> <mi>μ</mi> <mover accent="true"> <mi>a</mi> <mo>→</mo> </mover> </mrow> </math>               …(2)From (1) and (2)  <math> <mrow> <mover accent="true"> <mi>b</mi> <mo>→</mo> </mover> <mo>+</mo> <mn>6</mn> <mover accent="true"> <mi>c</mi> <mo>→</mo> </mover> <mo>=</mo> <mi>μ</mi> <mrow> <mo>(</mo> <mrow> <mi>λ</mi> <mover accent="true"> <mi>c</mi> <mo>→</mo> </mover> <mo>−</mo> <mn>5</mn> <mover accent="true"> <mi>b</mi> <mo>→</mo> </mover> </mrow> <mo>)</mo> </mrow> </mrow> </math>          -> <math> <mrow> <mrow> <mo>(</mo> <mrow> <mn>1</mn> <mo>+</mo> <mn>5</mn> <mi>μ</mi> </mrow> <mo>)</mo> </mrow> <mover accent="true"> <mi>b</mi> <mo>→</mo> </mover> <mo>+</mo> <mrow> <mo>(</mo> <mrow> <mn>6</mn> <mo>−</mo> <mi>λ</mi> <mi>μ</mi> </mrow> <mo>)</mo> </mrow> <mover accent="true"> <mi>c</mi> <mo>→</mo> </mover> <mo>=</mo> <mn>0</mn> </mrow> </math> <math> <mrow> <mo>?</mo> <mtext> </mtext> <mover accent="true"> <mi>b</mi> <mo>→</mo> </mover> </mrow> </math> and <math> <mrow> <mover accent="true"> <mi>c</mi> <mo>→</mo> </mover> </mrow> </math>  are non-collinear-> 1 + 5m = 0 <math> <mrow> <mi>μ</mi> <mo>=</mo> <mfrac> <mrow> <mo>−</mo> <mn>1</mn> </mrow> <mrow> <mn>5</mn> </mrow> </mfrac> </mrow> </math>  and 6 – lm = 0 Þ lm = 6-> l = – 30Now, <math> <mrow> <mover accent="true"> <mi>b</mi> <mo>→</mo> </mover> <mo>=</mo> <mn>6</mn> <mover accent="true"> <mi>c</mi> <mo>→</mo> </mover> <mo>=</mo> <mfrac> <mrow> <mo>−</mo> <mn>1</mn> </mrow> <mrow> <mn>5</mn> </mrow> </mfrac> <mover accent="true"> <mi>a</mi> <mo>→</mo> </mover> </mrow> </math> <math> <mrow> <mn>5</mn> <mover accent="true"> <mi>b</mi> <mo>→</mo> </mover> <mo>+</mo> <mn>3</mn> <mn>0</mn> <mover accent="true"> <mi>c</mi> <mo>→</mo> </mover> <mo>=</mo> <mo>−</mo> <mover accent="true"> <mi>a</mi> <mo>→</mo> </mover> </mrow> </math> <math> <mrow> <mrow> <mrow> <mtable> <mtr columnalign="center"> <mtd columnalign="center"> <mover accent="true"> <mi>a</mi> <mo>→</mo> </mover> <mo>+</mo> <mn>5</mn> <mover accent="true"> <mi>b</mi> <mo>→</mo> </mover> <mo>+</mo> <mn>3</mn> <mn>0</mn> <mover accent="true"> <mi>c</mi> <mo>→</mo> </mover> <mo>=</mo> <mn>0</mn> </mtd> <mtd columnalign="center"> <mover accent="true"> <mi>a</mi> <mo>→</mo> </mover> <mo>+</mo> <mi>α</mi> <mover accent="true"> <mi>b</mi> <mo>→</mo> </mover> <mo>+</mo> <mi>β</mi> <mover accent="true"> <mi>c</mi> <mo>→</mo> </mover> <mo>=</mo> <mn>0</mn> </mtd> </mtr> </mtable> </mrow> <mo>]</mo> </mrow> </mrow> </math> On comparingα = 5, β = 30 ⇒ α + β = 35

0 Upvotes 0 Downvotes

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
682k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

If $\vec{a}, \vec{b}, \vec{c}$ are non-zero and $\vec{b}$ and $\vec{c}$ are non-collinear. $\vec{a} + 5 \vec{b}$ is collinear with $\vec{c}$ and $\vec{b} + 6 \vec{c}$ is collinear with $\vec{a}$ . If $\vec{a} + α \vec{b} + β \vec{b} = 0$ , then find α + β.

1 Answer

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

If a → , b → , c → are non-zero and b → and c → are non-collinear. a → + 5 b → is collinear with c → and b → + 6 c → is collinear with a → . If a → + α b → + β b → = 0 , then find α + β.

1 Answer

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

If $\vec{a}, \vec{b}, \vec{c}$ are non-zero and $\vec{b}$ and $\vec{c}$ are non-collinear. $\vec{a} + 5 \vec{b}$ is collinear with $\vec{c}$ and $\vec{b} + 6 \vec{c}$ is collinear with $\vec{a}$ . If $\vec{a} + α \vec{b} + β \vec{b} = 0$ , then find α + β.