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Mathematics Question on Vector Algebra

Let a⃗=αi^+j^+βk^ and b⃗=3i^+5j^+4k^ be two vectors, such that a⃗×b⃗=−i^+9j^+12k^.\begin{array}{l}\text{Let}~\vec{a} = \alpha \hat{i} + \hat{j} + \beta \hat{k}~\text{and}~ \vec{b} = 3 \hat{i} + 5\hat{j} + 4 \hat{k}~ \text{be two vectors, such that }\vec{a} \times \vec{b} = -\hat{i} + 9\hat{j} + 12 \hat{k}.\end{array}
 Then the projection of b⃗−2a⃗ on b⃗+a⃗ is equal to\begin{array}{l}~\text{Then the projection of } \vec{b}-2\vec{a} ~\text{on}~ \vec{b}+ \vec{a}~\text {is equal to}\end{array}

A

2

B

395\frac{39}{5}

C

9

D

465\frac{46}{5}

Answer

465\frac{46}{5}

Explanation

Solution

a⃗=αi^+j^+βk^,b⃗=3i^–5j^+4k^\begin{array}{l}\vec{a}=\alpha \hat{i} + \hat{j}+ \beta \hat{k}, \vec{b} = 3\hat{i} – 5\hat{j}+ 4 \hat{k}\end{array}
a⃗×b⃗=−i^+9j^+12k^\begin{array}{l}\vec{a}\times \vec{b} = -\hat{i} + 9\hat{j}+ 12 \hat{k}\end{array}
∣i^j^k^α1β\3−54∣=−i^+9j^+12k^\begin{array}{l}\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\\\\alpha & 1 & \beta \\\3 & -5 & 4 \\\\\end{vmatrix}=-\hat{i}+9\hat{j}+12\hat{k}\end{array}
4+5β=−1⇒β=−1 −5α–3=12⇒α=−3\begin{array}{l}4 + 5\beta = -1 \Rightarrow \beta = -1\\\ -5\alpha – 3 = 12 \Rightarrow \alpha = -3\end{array}
b⃗−2a⃗=3i^−5j^+4k^−2(−3i^+j^−k^)\begin{array}{l}\vec{b}-2\vec{a}=3\hat{i}-5\hat{j}+4\hat{k}-2(-3\hat{i} + \hat{j}-\hat{k})\end{array}
b⃗−2a⃗=9i^−7j^+6k^\begin{array}{l}\vec{b}-2\vec{a}=9\hat{i}-7\hat{j}+6\hat{k}\end{array}
b⃗+a⃗=(3i^−5j^+4k^)+(−3i^+j^−k^)\begin{array}{l}\vec{b}+ \vec{a}=(3\hat{i}-5\hat{j}+4\hat{k}) + (-3\hat{i}+\hat{j}-\hat{k})\end{array}
b⃗+a⃗=−4j^+3k^\begin{array}{l}\vec{b}+ \vec{a}=-4\hat{j}+3\hat{k} \end{array}
Projection of b⃗−2a⃗ on b⃗+a⃗=(b⃗−2a⃗)⋅(b⃗+a⃗)∣b⃗+a⃗∣\begin{array}{l}\text{Projection of }\vec{b}- 2\vec{a} \text{ on } \vec{b} + \vec{a} = \frac{(\vec{b}-2\vec{a})\cdot(\vec{b} + \vec{a})}{|\vec{b} + \vec{a}|}\end{array}
=28+185=465\begin{array}{l}=\frac{28+18}{5}=\frac{46}{5}\end{array}