If a is a nonzero vector such that its projections on the vectors 2i - j + 2k, i + 2j - 2k and k are equal, then a unit vector along a is :

The correct answer is Option 4. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q11 – JEE-MAIN 2025

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Home Maths Vector AlgebraQuestion #516

\vec{a}

is a nonzero 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 :

Clue 1 of 3: Conceptual Principle

The projection of a vector $\vec{a}$ onto another vector $\vec{b}$ is given by the formula $\frac{\vec{a} \cdot \vec{b}}{| \vec{b} |}$ .

🥷

Ninja Strategy•Component Proportionality Check

After deriving the relationships $x = \frac{7}{5} z$ and $y = \frac{9}{5} z$ , quickly check which option's components satisfy these proportionalities and sign consistency.

🧠 Full Solution Path

● Video Walkthrough

Step 1: Define Vectors and Projection Formula✦ Active

Let the unknown vector be $\vec{a} = x \hat{i} + y \hat{j} + z \hat{k}$ . The given vectors are $\vec{b_{1}} = 2 \hat{i} - \hat{j} + 2 \hat{k}$ , $\vec{b_{2}} = \hat{i} + 2 \hat{j} - 2 \hat{k}$ , and $\vec{b_{3}} = \hat{k}$ . The projection of $\vec{a}$ on $\vec{b}$ is $P = \frac{\vec{a} \cdot \vec{b}}{| \vec{b} |}$ . We calculate the magnitudes of the given vectors:

| \vec{b_{1}} | = \sqrt{2^{2} + (- 1)^{2} + 2^{2}} = \sqrt{9} = 3

| \vec{b_{2}} | = \sqrt{1^{2} + 2^{2} + (- 2)^{2}} = \sqrt{9} = 3

| \vec{b_{3}} | = \sqrt{0^{2} + 0^{2} + 1^{2}} = \sqrt{1} = 1

Step 2: Formulate and Solve Equations for Components○ Expand

The projections are equal. Let the common projection value be $k$ . This gives us:

\frac{\vec{a} \cdot \vec{b_{1}}}{| \vec{b_{1}} |} = \frac{2 x - y + 2 z}{3} = k ⟹ 2 x - y + 2 z = 3 k (1)

\frac{\vec{a} \cdot \vec{b_{2}}}{| \vec{b_{2}} |} = \frac{x + 2 y - 2 z}{3} = k ⟹ x + 2 y - 2 z = 3 k (2)

\frac{\vec{a} \cdot \vec{b_{3}}}{| \vec{b_{3}} |} = \frac{z}{1} = k ⟹ z = k (3)

Substitute $z = k$ into equations (1) and (2):

2 x - y + 2 k = 3 k ⟹ 2 x - y = k (4)

x + 2 y - 2 k = 3 k ⟹ x + 2 y = 5 k (5)

Multiply equation (4) by 2: $4 x - 2 y = 2 k$ . Add this to equation (5): $(4 x - 2 y) + (x + 2 y) = 2 k + 5 k ⟹ 5 x = 7 k ⟹ x = \frac{7}{5} k$ . Substitute $x$ back into equation (4): $2 (\frac{7}{5} k) - y = k ⟹ \frac{14}{5} k - y = k ⟹ y = \frac{14}{5} k - k = \frac{9}{5} k$ . Thus, the components of $\vec{a}$ are $x = \frac{7}{5} k$ , $y = \frac{9}{5} k$ , and $z = k$ .

💡 Teacher's Secret Hint

Ensure careful algebraic manipulation when solving the system of equations.

Step 3: Determine the Unit Vector○ Expand

We can choose any non-zero value for $k$ . For simplicity, let $k = 5$ . Then $\vec{a} = 7 \hat{i} + 9 \hat{j} + 5 \hat{k}$ . Now, we find the magnitude of $\vec{a}$ :

| \vec{a} | = \sqrt{7^{2} + 9^{2} + 5^{2}} = \sqrt{49 + 81 + 25} = \sqrt{155}

The unit vector along $\vec{a}$ is $\hat{a} = \frac{\vec{a}}{| \vec{a} |}$ :

\hat{a} = \frac{1}{\sqrt{155}} (7 \hat{i} + 9 \hat{j} + 5 \hat{k})

This matches option 4.

💡 Teacher's Secret Hint

Remember that a unit vector has a magnitude of 1. The choice of $k$ only affects the magnitude of $\vec{a}$ , not its direction.

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Maths Question 11 – JEE-MAIN 2025

Clue 1 of 3: Conceptual Principle

🧠 Full Solution Path