Let a→=αi^+3j^−k^, b→=3i^−βj^+4k^ and c→=i^+2j^−2k^ where α,β∈R, be three vectors. If the projection of a→ on c→ is 103 and b→×c→=−6i^+10j^+7k^, then the value of α+β is equal to :

a→⋅c→|c→|=103 ⇒α+6+21+4+4=103⇒α=2 and i^j^k^3−β412−2=−6i^+j^+k^ ⇒2β−8=−6⇒β=1⇒α+β=3

Let a→=αi^+3j^−k^, b→=3i^−βj^+4k^ and c→=i^+2j^−2k^ where α,β∈R, be three vectors. If the projection of a→ on c→ is 103 and b→×c→=−6i^+10j^+7k^, then the value of α+β is equal to :

a→⋅c→|c→|=103 ⇒α+6+21+4+4=103⇒α=2 and i^j^k^3−β412−2=−6i^+j^+k^ ⇒2β−8=−6⇒β=1⇒α+β=3

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Let $\vec{a} = α \hat{i} + 3 \hat{j} - \hat{k}, \vec{b} = 3 \hat{i} - β \hat{j} + 4 \hat{k}$ and $\vec{c} = \hat{i} + 2 \hat{j} - 2 \hat{k}$ where $α, β \in R$ , be three vectors. If the projection of $\vec{a}$ on $\vec{c}$ is $\frac{10}{3}$ and $\vec{b} \times \vec{c} = - 6 \hat{i} + 10 \hat{j} + 7 \hat{k}$ , then the value of $α + β$ is equal to :

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Detailed Solution

$\frac{\vec{a} \cdot \vec{c}}{| \vec{c} |} = \frac{10}{3} \Rightarrow \frac{|α + 6 + 2|}{\sqrt{1 + 4 + 4}} = \frac{10}{3} \Rightarrow α = 2 and |\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & - β & 4 \\ 1 & 2 & - 2 \end{array}| = - 6 \hat{i} + \hat{j} + \hat{k} \begin{aligned} \Rightarrow 2 β - 8 = - 6 \Rightarrow β = 1 \\ \Rightarrow α + β = 3 \end{aligned}$