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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 6 days ago

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alok kumar singh | Contributor-Level 10

6 days 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

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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.

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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.

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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.