Let a = a1i + a2j + a3k, ai 0, i=1, 2, 3 be a vector which makes equal angles with the coordinate

Q: Why can't we just use a rotation matrix to find $\vec{b}$?

While you could technically use a rotation matrix, it is often overkill for JEE problems. The coplanarity condition provides a much more elegant algebraic path. By expressing $\vec{b}$ as a linear combination of $\vec{a}$ and $\hat{i}$, you bypass complex trigonometric matrices and rely on the fundamental properties of vectors, which is exactly what the examiners want to see.

Q: Why does the projection result in $\pm \sqrt{2}$?

Remember that the projection of one vector onto another is a scalar quantity, not a magnitude. It can be positive, negative, or zero depending on the angle between the vectors. Since our scalar $\alpha$ can be $\pm \frac{1}{\sqrt{2}}$, the projection naturally carries that sign. The question asks for the projection, and the result is consistent with the geometry of the rotation.

Q: How do we know $\vec{b}$ is in the plane of $\vec{a}$ and $\hat{i}$?

The problem explicitly states that $\vec{a}$, $\vec{b}$, and the x-axis are coplanar. This is a geometric constraint given to us. Whenever you see 'coplanar' in a vector problem, immediately think: 'linear combination'. It is the most powerful tool in your arsenal for handling coplanarity.

Question

Let $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$, $a_i > 0, i=1, 2, 3$ be a vector which makes equal angles with the coordinate axes $OX, OY$ and $OZ$. Also, let the projection of $\vec{a}$ on the vector $3\hat{i} + 4\hat{j}$ be 7. Let $\vec{b}$ be a vector obtained by rotating $\vec{a}$ with $90^\circ$. If $\vec{a}, \vec{b}$ and x-axis are coplanar, then projection of a vector $\vec{b}$ on $3\hat{i} + 4\hat{j}$ is equal to

Options:

(A) $\sqrt{7}$

(B) $\sqrt{2}$

(C) 2

(D) 7

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in the center of a perfectly cubic room. You are asked to draw a vector $\vec{a}$ that makes equal angles with the walls, the floor, and the ceiling.

When a vector makes equal angles with the coordinate axes $OX$, $OY$, and $OZ$, its direction cosines $l$, $m$, and $n$ must be equal. Since the sum of the squares of the direction cosines is always unity, we have $l^2 + m^2 + n^2 = 1$, which simplifies to $3l^2 = 1$.

Thus, $l = m = n = \frac{1}{\sqrt{3}}$. This tells us that the direction of $\vec{a}$ is simply the direction of the vector $\hat{i} + \hat{j} + \hat{k}$. We can write $\vec{a} = \lambda(\hat{i} + \hat{j} + \hat{k})$ for some positive scalar $\lambda$.

### The Projection Constraint: Finding the Magnitude

Now, we are given a target vector $\vec{v} = 3\hat{i} + 4\hat{j}$. The problem states that the projection of $\vec{a}$ onto $\vec{v}$ is $7$.

Recall the projection formula: the projection of $\vec{a}$ on $\vec{v}$ is $\frac{\vec{a} \cdot \vec{v}}{|\vec{v}|}$. The magnitude of $\vec{v}$ is $|\vec{v}| = \sqrt{3^2 + 4^2} = 5$.

Substituting our expression for $\vec{a}$, we get:

$$ \frac{\lambda(\hat{i} + \hat{j} + \hat{k}) \cdot (3\hat{i} + 4\hat{j})}{5} = 7 $$

This simplifies to $\frac{\lambda(3 + 4)}{5} = 7$, which gives us $\frac{7\lambda}{5} = 7$. Solving this, we find $\lambda = 5$. Our vector $\vec{a}$ is now fully revealed: $\vec{a} = 5\hat{i} + 5\hat{j} + 5\hat{k}$.

### The Dance of Rotation and Coplanarity

Next, we introduce vector $\vec{b}$, which is $\vec{a}$ rotated by $90^\circ$. Rotation preserves the length of the vector. Thus, $|\vec{b}| = |\vec{a}| = \sqrt{5^2 + 5^2 + 5^2} = 5\sqrt{3}$.

Furthermore, a $90^\circ$ rotation implies that $\vec{a}$ and $\vec{b}$ are orthogonal, meaning $\vec{a} \cdot \vec{b} = 0$. The problem states that $\vec{a}$, $\vec{b}$, and the x-axis (represented by $\hat{i}$) are coplanar.

In the language of linear algebra, this means $\vec{b}$ can be expressed as a linear combination of $\vec{a}$ and $\hat{i}$. We write $\vec{b} = \alpha\vec{a} + \beta\hat{i}$. Substituting $\vec{a}$, we get:

$$ \vec{b} = \alpha(5\hat{i} + 5\hat{j} + 5\hat{k}) + \beta\hat{i} = (5\alpha + \beta)\hat{i} + 5\alpha\hat{j} + 5\alpha\hat{k} $$

### Solving the Algebraic Puzzle

We have two unknowns, $\alpha$ and $\beta$. We use our two conditions: orthogonality and magnitude. First, $\vec{a} \cdot \vec{b} = 0$:

$$ (5\hat{i} + 5\hat{j} + 5\hat{k}) \cdot ((5\alpha + \beta)\hat{i} + 5\alpha\hat{j} + 5\alpha\hat{k}) = 0 $$

This expands to $5(5\alpha + \beta) + 5(5\alpha) + 5(5\alpha) = 0$, which simplifies to $75\alpha + 5\beta = 0$. Thus, $\beta = -15\alpha$.

Now, we substitute this back into our expression for $\vec{b}$:

$$ \vec{b} = (5\alpha - 15\alpha)\hat{i} + 5\alpha\hat{j} + 5\alpha\hat{k} = -10\alpha\hat{i} + 5\alpha\hat{j} + 5\alpha\hat{k} $$

Finally, we use the magnitude condition $|\vec{b}|^2 = 75$:

$$ (-10\alpha)^2 + (5\alpha)^2 + (5\alpha)^2 = 75 \Rightarrow 150\alpha^2 = 75 \Rightarrow \alpha^2 = \frac{1}{2} $$

Thus, $\alpha = \pm \frac{1}{\sqrt{2}}$.

### The Final Projection

The final step is to find the projection of $\vec{b}$ on $3\hat{i} + 4\hat{j}$. Using the projection formula again:

$$ \frac{\vec{b} \cdot (3\hat{i} + 4\hat{j})}{5} = \frac{(-10\alpha)(3) + (5\alpha)(4)}{5} = \frac{-30\alpha + 20\alpha}{5} = -2\alpha $$

Substituting $\alpha = \pm \frac{1}{\sqrt{2}}$, the projection is $\mp \sqrt{2}$. The magnitude of this projection is **$\sqrt{2}$**.

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Visualized Solution

Direction of Vector $a$

Projection Formula

Finding $λ$

Properties of Vector $b$

Coplanarity Condition

Expressing $b$

Using Orthogonality

Refining $b$

Finding $α$

Final Projection

Conclusion

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

The Projection Constraint

The Dance of Rotation and Coplanarity

Solving the Algebraic Puzzle

The Final Projection

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Visualized Solution

Direction of Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Projection Formula

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

Properties of Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b