Question

If \(\bar a\ and\ \bar b\) are position vectors of the points (a, 3, 0) and (1, 0, 0) respectively and if the angle between the vectors \(\bar a\ and\ \bar b\ is\ \frac{\pi}{4}\), then the value of a is equal to

• A. 1
• B. 2
• C. 3
• D. 4
• E. 5

Answer 1

The Correct Option is C

Let \( \mathbf{a} = (\alpha, 3, 0) \) and \( \mathbf{b} = (1, 0, 0) \). The dot product formula is given by: \[ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta \] where \( \theta \) is the angle between the vectors. We are given that \( \theta = \frac{\pi}{4} \), so \( \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} \). ### Step 1: Calculate the dot product \( \mathbf{a} \cdot \mathbf{b} \) \[ \mathbf{a} \cdot \mathbf{b} = \alpha \cdot 1 + 3 \cdot 0 + 0 \cdot 0 = \alpha \] ### Step 2: Calculate the magnitudes of \( \mathbf{a} \) and \( \mathbf{b} \) \[ |\mathbf{a}| = \sqrt{\alpha^2 + 3^2 + 0^2} = \sqrt{\alpha^2 + 9} \] \[ |\mathbf{b}| = \sqrt{1^2 + 0^2 + 0^2} = 1 \] ### Step 3: Use the dot product formula Substitute the values into the dot product formula: \[ \alpha = |\mathbf{a}| \cdot 1 \cdot \cos \frac{\pi}{4} \] \[ \alpha = \sqrt{\alpha^2 + 9} \cdot \frac{1}{\sqrt{2}} \] Multiply both sides by \( \sqrt{2} \): \[ \alpha \sqrt{2} = \sqrt{\alpha^2 + 9} \] ### Step 4: Square both sides \[ 2\alpha^2 = \alpha^2 + 9 \] Simplify the equation: \[ \alpha^2 = 9 \] Thus: \[ \alpha = 3 \]

The correct option is (C) : \(3\)

Answer 2

We are given that the position vector of point (a, 3, 0) is \(\bar{a} = a\hat{i} + 3\hat{j} + 0\hat{k}\) and the position vector of point (1, 0, 0) is \(\bar{b} = 1\hat{i} + 0\hat{j} + 0\hat{k}\).

The angle between the vectors \(\bar{a}\) and \(\bar{b}\) is \(\frac{\pi}{4}\).

We know that \(\bar{a} \cdot \bar{b} = |\bar{a}||\bar{b}|\cos\theta\), where \(\theta\) is the angle between \(\bar{a}\) and \(\bar{b}\).

First, let's find \(\bar{a} \cdot \bar{b}\):

\(\bar{a} \cdot \bar{b} = (a\hat{i} + 3\hat{j} + 0\hat{k}) \cdot (1\hat{i} + 0\hat{j} + 0\hat{k}) = a(1) + 3(0) + 0(0) = a\)

Now, let's find \(|\bar{a}|\) and \(|\bar{b}|\):

\(|\bar{a}| = \sqrt{a^2 + 3^2 + 0^2} = \sqrt{a^2 + 9}\)

\(|\bar{b}| = \sqrt{1^2 + 0^2 + 0^2} = \sqrt{1} = 1\)

We are given that \(\theta = \frac{\pi}{4}\), so \(\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\).

Therefore, we have:

\(a = \sqrt{a^2 + 9} \cdot 1 \cdot \frac{\sqrt{2}}{2}\)

\(a = \sqrt{a^2 + 9} \cdot \frac{\sqrt{2}}{2}\)

Multiplying by \(\frac{2}{\sqrt{2}}\) we get \(\sqrt{2} a = \sqrt{a^2 + 9}\).

Squaring both sides, we have:

\(2a^2 = a^2 + 9\)

\(a^2 = 9\)

\(a = \pm 3\)

Since, we squared it lets check it

Substituting into original equation

\(a = \sqrt{a^2 + 9} \cdot \frac{\sqrt{2}}{2}\)

If a = 3

\(3 = \sqrt{3^2 + 9} \cdot \frac{\sqrt{2}}{2} = \sqrt{18} \cdot \frac{\sqrt{2}}{2} = 3\sqrt{2} \frac{\sqrt{2}}{2} = 3\)

If a = -3

\(-3 = \sqrt{(-3)^2 + 9} \cdot \frac{\sqrt{2}}{2} = \sqrt{18} \cdot \frac{\sqrt{2}}{2} = 3\sqrt{2} \frac{\sqrt{2}}{2} = 3\)

Since we are given that b = (1, 0, 0) if we use a = -3 the angle between vectors will be obtuse since inner product will be negative violating that \(\theta = \frac{\pi}{4}\) Thus a = 3;