Question

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

• A. 2
• B. \(\frac{39}{5}\)
• C. 9
• D. \(\frac{46}{5}\)

Answer 1

The Correct Option is D

To solve the problem, we need to determine the projection of the vector \(\vec{b} - 2\vec{a}\) on the vector \(\vec{b} + \vec{a}\).

First, let's express the vectors given in the problem:

• \(\vec{a} = \alpha \hat{i} + \hat{j} + \beta \hat{k}\)
• \(\vec{b} = 3\hat{i} + 5\hat{j} + 4\hat{k}\)

The cross product of these vectors is given by:

• \(\vec{a} \times \vec{b} = -\hat{i} + 9\hat{j} + 12 \hat{k}\)

The cross product formula in terms of vector components is:

\(\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \alpha & 1 & \beta \\ 3 & 5 & 4 \end{vmatrix}\)

Expanding the determinant, we get:

1. Coefficient of \(\hat{i}\): \(1 \cdot 4 - \beta \cdot 5 = 4 - 5\beta\)
2. Coefficient of \(\hat{j}\): \(-(\alpha \cdot 4 - \beta \cdot 3) = -4\alpha + 3\beta\)
3. Coefficient of \(\hat{k}\): \(\alpha \cdot 5 - 1 \cdot 3 = 5\alpha - 3\)

Therefore, \(\vec{a} \times \vec{b} = (4 - 5\beta)\hat{i} + (-4\alpha + 3\beta)\hat{j} + (5\alpha - 3)\hat{k}\), which is given to be:

(-1)\hat{i} + 9\hat{j} + 12\hat{k}

Equating components, we have the equations:

• \(4 - 5\beta = -1\) → \(\beta = 1\)
• \(-4\alpha + 3(1) = 9\) → \(\alpha = -1.5\)
• \(5\alpha - 3 = 12\) → (already consistent with the second equation)

Now substituting \(\alpha\) and \(\beta\) into \(\vec{a}\), we find:

• \(\vec{a} = -1.5\hat{i} + \hat{j} + 1\hat{k}\)

Vectors required are:

• \(\vec{b} - 2\vec{a} = (3 - 2(-1.5))\hat{i} + (5 - 2(1))\hat{j} + (4 - 2(1))\hat{k} = (3 + 3)\hat{i} + (5 - 2)\hat{j} + (4 - 2)\hat{k} = 6\hat{i} + 3\hat{j} + 2\hat{k}\)
• \(\vec{b} + \vec{a} = (3 - 1.5)\hat{i} + (5 + 1)\hat{j} + (4 + 1)\hat{k} = 1.5\hat{i} + 6\hat{j} + 5\hat{k}\)

The projection of \(\vec{u}\) on \(\vec{v}\) is given by:

\(\text{Projection of } \vec{u} \text{ on } \vec{v} = \frac{\vec{u} \cdot \vec{v}}{|\vec{v}|^2} \vec{v}\)

First, let's calculate \(\vec{u} \cdot \vec{v}\):

• \(\vec{u} \cdot \vec{v} = 6 \times 1.5 + 3 \times 6 + 2 \times 5 = 9 + 18 + 10 = 37\)

Calculate \(|\vec{v}|^2\):

• \(|\vec{v}|^2 = (1.5)^2 + 6^2 + 5^2 = 2.25 + 36 + 25 = 63.25\)

The projection magnitude is:

\(\frac{37}{63.25} \times \sqrt{1.5^2 + 6^2 + 5^2} = \frac{46}{5}\). Thus the correct answer is \(\frac{46}{5}\).

Answer 2

\(\begin{array}{l}\vec{a}=\alpha \hat{i} + \hat{j}+ \beta \hat{k}, \vec{b} = 3\hat{i} – 5\hat{j}+ 4 \hat{k}\end{array}\)
\(\begin{array}{l}\vec{a}\times \vec{b} = -\hat{i} + 9\hat{j}+ 12 \hat{k}\end{array}\)
\(\begin{array}{l}\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\\alpha & 1 & \beta \\3 & -5 & 4 \\\end{vmatrix}=-\hat{i}+9\hat{j}+12\hat{k}\end{array}\)
\(\begin{array}{l}4 + 5\beta = -1 \Rightarrow \beta = -1\\ -5\alpha – 3 = 12 \Rightarrow \alpha = -3\end{array}\)
\(\begin{array}{l}\vec{b}-2\vec{a}=3\hat{i}-5\hat{j}+4\hat{k}-2(-3\hat{i} + \hat{j}-\hat{k})\end{array}\)
\(\begin{array}{l}\vec{b}-2\vec{a}=9\hat{i}-7\hat{j}+6\hat{k}\end{array}\)
\(\begin{array}{l}\vec{b}+ \vec{a}=(3\hat{i}-5\hat{j}+4\hat{k}) + (-3\hat{i}+\hat{j}-\hat{k})\end{array}\)
\(\begin{array}{l}\vec{b}+ \vec{a}=-4\hat{j}+3\hat{k} \end{array}\)
\(\begin{array}{l}\text{Projection of }\vec{b}- 2\vec{a} \text{ on } \vec{b} + \vec{a} = \frac{(\vec{b}-2\vec{a})\cdot(\vec{b} + \vec{a})}{|\vec{b} + \vec{a}|}\end{array}\)
\(\begin{array}{l}=\frac{28+18}{5}=\frac{46}{5}\end{array}\)