Question

Let \(\vec a\)=\(\hat i\)−\(\hat j\)+2\(\hat k\) and \(\vec b\) be a vector such that 
\(\vec a\)×\(\vec b\)=2\(\hat i\)−\(\hat k\) and \(\vec a\)⋅\(\vec b\)=3. 
Then the projection of  \(\vec b\) on the vector \(\vec a\)−\(\vec b\) is :

• A. \(\frac{2}{\sqrt{21}}\)
• B. \(\frac{2\sqrt3}{7}\)
• C. \(\frac{2}{\frac{3\sqrt7}{3}}\)
• D. \(\frac{2}{3}\)

Answer 1

The Correct Option is A

To find the projection of \(\vec{b}\) on \(\vec{a} - \vec{b}\), we'll follow these steps:

1. Determine the vector \(\vec{a} - \vec{b}\). 
2. Use the formula for the projection of a vector: \[ \text{Projection of } \vec{u} \text{ on } \vec{v} = \frac{\vec{u} \cdot \vec{v}}{\|\vec{v}\|^2} \vec{v} \]

Given, \( \vec{a} = \hat{i} - \hat{j} + 2\hat{k} \)

And, \(\vec{a} \times \vec{b} = 2\hat{i} - \hat{k}\)

Let's express \(\vec{b}\) as \(x\hat{i} + y\hat{j} + z\hat{k}\).

The cross product \( \vec{a} \times \vec{b} \) is given by the determinant:

\(\hat{i}\)	\(\hat{j}\)	\(\hat{k}\)
1	-1	2
x	y	z

This results in:

\[ \vec{a} \times \vec{b} = \left((-1 \cdot z - 2 \cdot y)\hat{i} - (2 \cdot x - 1 \cdot z)\hat{j} + (1 \cdot y - (-1) \cdot x)\hat{k}\right) \] \[= (-z - 2y)\hat{i} + (-2x + z)\hat{j} + (y + x)\hat{k}\]

Equating to \(2\hat{i} - \hat{k}\), we get:

• \(-z - 2y = 2\) ... (i)
• \(y + x = -1\) ... (ii)

Also given is \(\vec{a} \cdot \vec{b} = 3\):

\[ \vec{a} \cdot \vec{b} = 1 \cdot x + (-1) \cdot y + 2 \cdot z = x - y + 2z = 3 \text{... (iii)} \]

Solving equations (i), (ii), and (iii) simultaneously:

• From (ii), express \(y = -1 - x\).
• Substitute in (i): \(-z - 2(-1-x) = 2\)
• \(-z + 2 + 2x = 2\)
• \(z = 2x\) ... (iv)
• Substitute y and z in (iii):
• \(x - (-1-x) + 2(2x) = 3\)
• \(x + 1 + x + 4x = 3\)
• \(6x = 2\)
• \(x = \frac{1}{3}\)
• \(y = -1 - x = -\frac{4}{3}\)
• \(z = 2x = \frac{2}{3}\)

So, \(\vec{b} = \frac{1}{3}\hat{i} - \frac{4}{3}\hat{j} + \frac{2}{3}\hat{k}\).

Now calculate \(\vec{a} - \vec{b} = \left(\hat{i} - \hat{j} + 2\hat{k}\right) - \left(\frac{1}{3}\hat{i} - \frac{4}{3}\hat{j} + \frac{2}{3}\hat{k}\right)\).

\[ = \left(\hat{i} - \frac{1}{3}\hat{i}\right) - \left(\hat{j} + \frac{4}{3}\hat{j}\right) + \left(2\hat{k} - \frac{2}{3}\hat{k}\right) \\ = \frac{2}{3}\hat{i} - \frac{7}{3}\hat{j} + \frac{4}{3}\hat{k} \]

Now, let's find the projection of \(\vec{b}\) on \(\vec{a} - \vec{b}\):

First calculate the dot product \(\vec{b} \cdot (\vec{a} - \vec{b})\):

\[ = \left(\frac{1}{3}\cdot\frac{2}{3}\right) + \left(-\frac{4}{3}\cdot\left(-\frac{7}{3}\right)\right) + \left(\frac{2}{3}\cdot\frac{4}{3}\right) \] \[ = \frac{2}{9} + \frac{28}{9} + \frac{8}{9} = \frac{38}{9} \]

Magnitude of \(\vec{a} - \vec{b}\):

\[ = \sqrt{\left(\frac{2}{3}\right)^2 + \left(\frac{7}{3}\right)^2 + \left(\frac{4}{3}\right)^2} \] \[ = \sqrt{\frac{4}{9} + \frac{49}{9} + \frac{16}{9}} \] \[ = \sqrt{\frac{69}{9}} = \sqrt{\frac{23}{3}} \]

Thus, \[ \text{Projection of } \vec{b} \text{ on } \vec{a} - \vec{b} = \frac{\frac{38}{9}}{\frac{23}{3}} = \frac{38}{9} \times \frac{3}{23} = \frac{2}{\sqrt{21}} \]

Therefore, the projection of \(\vec{b}\) on \(\vec{a} - \vec{b}\) is \(\boxed{\frac{2}{\sqrt{21}}}\).

Answer 2

\[ \vec a = \hat i - \hat j + 2 \hat k \] 

\[ \vec a \times \vec b = 2 \hat i - \hat k \]

\[ \vec a \cdot \vec b = 3 | \vec a \times \vec b |^2 + | \vec a \cdot \vec b |^2 \]

\[ = | \vec a |^2 \cdot | \vec b |^2 \]

\[ \Rightarrow 5 + 9 = 6 | \vec b |^2 \]

\[ \Rightarrow | \vec b |^2 = \frac{7}{3} \]

\[ | \vec a - \vec b | = \sqrt{| \vec a |^2 + | \vec b |^2 - 2 \vec a \cdot \vec b} = \frac{\sqrt{7}}{3} \]

Projection of \( \vec b \) on \( \vec a - \vec b \) is:

\[ \frac{\vec b (\vec a - \vec b)}{| \vec a - \vec b |} = \frac{2}{\sqrt{21}} \]