Let vec{a}, vec{b} and vec{c} be three non-ze | vedclass

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(A) Given that $\vec{a}+5\vec{b}$ is collinear with $\vec{c}$,there exists a scalar $\lambda$ such that $\vec{a}+5\vec{b} = \lambda\vec{c} \implies \v

Vedclass · Accepted Answer

$35$

Vedclass · Answer

(A) Given that $\vec{a}+5\vec{b}$ is collinear with $\vec{c}$,there exists a scalar $\lambda$ such that $\vec{a}+5\vec{b} = \lambda\vec{c} \implies \vec{a} = \lambda\vec{c} - 5\vec{b}$.Given that $\vec{b}+6\vec{c}$ is collinear with $\vec{a}$,there exists a scalar $\mu$ such that $\vec{b}+6\vec{c} = \mu\vec{a}$.Substituting the expression for $\vec{a}$ into the second equation:$\vec{b}+6\vec{c} = \mu(\lambda\vec{c} - 5\vec{b})$$\vec{b}+6\vec{c} = \mu\lambda\vec{c} - 5\mu\vec{b}$Rearranging the terms:$(1+5\mu)\vec{b} + (6-\mu\lambda)\vec{c} = \vec{0}$.Since $\vec{b}$ and $\vec{c}$ are non-collinear,their coefficients must be zero:$1+5\mu = 0 \implies \mu = -\frac{1}{5}$.$6-\mu\lambda = 0 \implies 6 - (-\frac{1}{5})\lambda = 0 \implies 6 + \frac{\lambda}{5} = 0 \implies \lambda = -30$.Now,substitute $\lambda$ back into the expression for $\vec{a}$:$\vec{a} = -30\vec{c} - 5\vec{b} \implies \vec{a} + 5\vec{b} + 30\vec{c} = \vec{0}$.Comparing this with $\vec{a} + \alpha\vec{b} + \beta\vec{c} = \vec{0}$,we get $\alpha = 5$ and $\beta = 30$.Therefore,$\alpha + \beta = 5 + 30 = 35$.

List-$I$	List-$II$
$(i)$ $\overrightarrow{a} \cdot \overrightarrow{b}$	$(A)$ $\overrightarrow{a} \cdot \overrightarrow{d}$
(ii) $\overrightarrow{b} \cdot \overrightarrow{c}$	$(B)$ $3$
(iii) $[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]$	$(C)$ $\overrightarrow{b} \cdot \overrightarrow{d}$
(iv) $\overrightarrow{b} \times \overrightarrow{c}$	$(D)$ $2\hat{i}-2\hat{k}$
	$(E)$ $2\hat{j}+2\hat{k}$
	$(F)$ $4$

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three non-zero vectors such that $\vec{b}$ and $\vec{c}$ are non-collinear. If $\vec{a}+5\vec{b}$ is collinear with $\vec{c}$,$\vec{b}+6\vec{c}$ is collinear with $\vec{a}$,and $\vec{a}+\alpha\vec{b}+\beta\vec{c}=\vec{0}$,then $\alpha+\beta$ is equal to

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