Question

If \( \vec{a} \) is a non-zero vector such that its projections on the vectors \( 2\hat{i} - \hat{j} + 2\hat{k},\ \hat{i} + 2\hat{j} - 2\hat{k} \), and \( \hat{k} \) are equal, then a unit vector along \( \vec{a} \) is:

• A. \( \frac{1}{\sqrt{155}} (7\hat{i} + 9\hat{j} + 5\hat{k}) \)
• B. \( \frac{1}{\sqrt{155}} (7\hat{i} + 9\hat{j} - 5\hat{k}) \)
• C. \( \frac{1}{\sqrt{155}} (7\hat{i} + 9\hat{j} + 5\hat{k}) \)
• D. \( \frac{1}{\sqrt{155}} (7\hat{i} + 9\hat{j} - 5\hat{k}) \)

Hint:

Use projection formulas and form simultaneous equations. Normalize the result to get a unit vector.

Answer 1

The Correct Option is C

We need to find a unit vector along a non-zero vector \( \vec{a} \), given that its projections on three other vectors, \( \vec{b}_1 = 2\hat{i} - \hat{j} + 2\hat{k} \), \( \vec{b}_2 = \hat{i} + 2\hat{j} - 2\hat{k} \), and \( \vec{b}_3 = \hat{k} \), are equal.

Concept Used:

The projection of a vector \( \vec{a} \) onto another non-zero vector \( \vec{b} \) is given by the formula:

\[ \text{Projection of } \vec{a} \text{ on } \vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \]

We will equate the projections of \( \vec{a} \) on the three given vectors to form a system of linear equations and solve for the components of \( \vec{a} \).

Step-by-Step Solution:

Step 1: Define the vector \( \vec{a} \) and the given vectors.

Let the unknown vector be \( \vec{a} = x\hat{i} + y\hat{j} + z\hat{k} \). The given vectors are:

\[ \vec{b}_1 = 2\hat{i} - \hat{j} + 2\hat{k} \] \[ \vec{b}_2 = \hat{i} + 2\hat{j} - 2\hat{k} \] \[ \vec{b}_3 = \hat{k} \]

Step 2: Calculate the magnitudes of the given vectors.

\[ |\vec{b}_1| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \] \[ |\vec{b}_2| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \] \[ |\vec{b}_3| = \sqrt{0^2 + 0^2 + 1^2} = 1 \]

Step 3: Set up the equality for the projections.

According to the problem, the projections are equal:

\[ \frac{\vec{a} \cdot \vec{b}_1}{|\vec{b}_1|} = \frac{\vec{a} \cdot \vec{b}_2}{|\vec{b}_2|} = \frac{\vec{a} \cdot \vec{b}_3}{|\vec{b}_3|} \]

Substituting the dot products and magnitudes:

\[ \frac{(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (2\hat{i} - \hat{j} + 2\hat{k})}{3} = \frac{(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (\hat{i} + 2\hat{j} - 2\hat{k})}{3} = \frac{(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (\hat{k})}{1} \] \[ \frac{2x - y + 2z}{3} = \frac{x + 2y - 2z}{3} = z \]

Step 4: Form and solve the system of linear equations.

From the equality, we can create two separate equations:

1) \( \frac{2x - y + 2z}{3} = z \implies 2x - y + 2z = 3z \implies 2x - y = z \quad \cdots(I) \)

2) \( \frac{x + 2y - 2z}{3} = z \implies x + 2y - 2z = 3z \implies x + 2y = 5z \quad \cdots(II) \)

To solve for \(x\) and \(y\) in terms of \(z\), we can multiply equation (I) by 2 and add it to equation (II):

\[ 2(2x - y) + (x + 2y) = 2(z) + 5z \] \[ 4x - 2y + x + 2y = 7z \] \[ 5x = 7z \implies x = \frac{7}{5}z \]

Now, substitute the value of \(x\) back into equation (I):

\[ 2\left(\frac{7}{5}z\right) - y = z \] \[ \frac{14}{5}z - y = z \implies y = \frac{14}{5}z - z = \frac{9}{5}z \]

Step 5: Determine the vector \( \vec{a} \) and find its unit vector.

Substitute the expressions for \(x\) and \(y\) back into \( \vec{a} \):

\[ \vec{a} = \left(\frac{7}{5}z\right)\hat{i} + \left(\frac{9}{5}z\right)\hat{j} + z\hat{k} \]

We can factor out \( \frac{z}{5} \):

\[ \vec{a} = \frac{z}{5} (7\hat{i} + 9\hat{j} + 5\hat{k}) \]

The unit vector \( \hat{a} \) is \( \frac{\vec{a}}{|\vec{a}|} \). The scalar factor \( \frac{z}{5} \) will cancel out, so the direction of \( \vec{a} \) is determined by the vector \( 7\hat{i} + 9\hat{j} + 5\hat{k} \).

\[ \hat{a} = \frac{7\hat{i} + 9\hat{j} + 5\hat{k}}{|7\hat{i} + 9\hat{j} + 5\hat{k}|} \]

Final Computation & Result:

First, calculate the magnitude of the direction vector:

\[ |7\hat{i} + 9\hat{j} + 5\hat{k}| = \sqrt{7^2 + 9^2 + 5^2} = \sqrt{49 + 81 + 25} = \sqrt{155} \]

Therefore, the unit vector along \( \vec{a} \) is:

\( \hat{a} = \frac{7\hat{i} + 9\hat{j} + 5\hat{k}}{\sqrt{155}} \)