The vector a = -i + 2j + k is rotated through a right angle, passing through the y-axis in its way a

Q: Wait, why do we use the Vector Triple Product? Can't we just rotate the vector using a rotation matrix?

You absolutely could use a rotation matrix, but that often involves complex trigonometry and finding the axis of rotation, which is overkill here. The Vector Triple Product $\vec{a} \times (\vec{a} \times \hat{j})$ is a powerful shortcut that keeps us strictly within the plane of interest. It is the elegant, geometric way to find a perpendicular vector in a specific plane without needing to calculate angles or rotation matrices.

Q: Why do we have to choose the negative sign for $\vec{b}$? How do I know which one is correct?

Great catch! The problem states the vector passes through the y-axis. This implies that as we rotate $\vec{a}$ towards the y-axis, the angle between the vector and the y-axis must decrease. By checking the dot product of our candidate vectors with $\hat{j}$, we ensure the one we choose moves towards the y-axis, satisfying the physical constraint of the path.

Q: Is the projection formula $\frac{\vec{R} \cdot \vec{c}}{|\vec{c}|}$ always valid?

Yes, it is! The projection of any vector $\vec{R}$ onto a vector $\vec{c}$ is defined as the scalar component of $\vec{R}$ in the direction of $\vec{c}$. The formula $\frac{\vec{R} \cdot \vec{c}}{|\vec{c}|}$ is the fundamental definition. Just ensure you calculate the dot product and the magnitude of the target vector $\vec{c}$ correctly, and you will never go wrong.

Question

The vector $\vec{a} = -\hat{i} + 2\hat{j} + \hat{k}$ is rotated through a right angle, passing through the y-axis in its way and the resulting vector is $\vec{b}$. Then the projection of $3\vec{a} + \sqrt{2}\vec{b}$ on $\vec{c} = 5\hat{i} + 4\hat{j} + 3\hat{k}$ is

Options:

(A) $3\sqrt{2}$

(B) 1

(C) $\sqrt{6}$

(D) $2\sqrt{3}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given the vector $\vec{a} = -\hat{i} + 2\hat{j} + \hat{k}$. We aim to rotate this vector by $90^\circ$ such that it sweeps through the $y$-axis.

The plane of rotation is defined by the vectors $\vec{a}$ and $\hat{j}$. We seek a vector $\vec{b}$ that is perpendicular to $\vec{a}$ and lies within this plane.

### The Vector Triple Product

To find a vector in the plane of $\vec{a}$ and $\hat{j}$ that is perpendicular to $\vec{a}$, we utilize the vector triple product:
$$\vec{V} = \vec{a} 	imes (\vec{a} 	imes \hat{j})$$

Applying the vector identity $\vec{a} 	imes (\vec{b} 	imes \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$, we expand the expression:
$$\vec{V} = (\vec{a} \cdot \hat{j})\vec{a} - |\vec{a}|^2\hat{j}$$

Calculating the necessary components:
1. $\vec{a} \cdot \hat{j} = 2$
2. $|\vec{a}|^2 = (-1)^2 + 2^2 + 1^2 = 6$

Substituting these values, we obtain:
$$\vec{V} = 2(-\hat{i} + 2\hat{j} + \hat{k}) - 6\hat{j} = -2\hat{i} - 2\hat{j} + 2\hat{k}$$

### Normalization and Constraints

We must normalize $\vec{V}$ to match the magnitude of $\vec{a}$, which is $|\vec{a}| = \sqrt{6}$. The magnitude of $\vec{V}$ is:
$$|\vec{V}| = \sqrt{(-2)^2 + (-2)^2 + 2^2} = \sqrt{12} = 2\sqrt{3}$$

Thus, the vector $\vec{b}$ is given by:
$$\vec{b} = \pm \frac{\sqrt{6}}{2\sqrt{3}} \vec{V} = \pm \frac{1}{\sqrt{2}} \vec{V}$$

Given the constraint that the path passes through the $y$-axis, the $y$-component must be positive. This yields:
$$\vec{b} = \sqrt{2}\hat{i} + \sqrt{2}\hat{j} - \sqrt{2}\hat{k}$$

### Final Calculation

We compute the linear combination $\vec{R} = 3\vec{a} + \sqrt{2}\vec{b}$:
$$\vec{R} = 3(-\hat{i} + 2\hat{j} + \hat{k}) + \sqrt{2}(\sqrt{2}\hat{i} + \sqrt{2}\hat{j} - \sqrt{2}\hat{k})$$
$$\vec{R} = (-3+2)\hat{i} + (6+2)\hat{j} + (3-2)\hat{k} = -\hat{i} + 8\hat{j} + \hat{k}$$

Finally, we find the projection of $\vec{R}$ onto $\vec{c} = 5\hat{i} + 4\hat{j} + 3\hat{k}$. The projection is defined as $\frac{\vec{R} \cdot \vec{c}}{|\vec{c}|}$:
$$\vec{R} \cdot \vec{c} = (-1)(5) + (8)(4) + (1)(3) = -5 + 32 + 3 = 30$$
$$|\vec{c}| = \sqrt{5^2 + 4^2 + 3^2} = \sqrt{25 + 16 + 9} = \sqrt{50} = 5\sqrt{2}$$

The final result is:
$$	ext{Projection} = \frac{30}{5\sqrt{2}} = \frac{6}{\sqrt{2}} = \mathbf{3\sqrt{2}}$$

Animated Solution for Mathematics - Vector Algebra: The vector $a = - \hat{i} + 2 \hat{j} + \hat{k}$ is rotated through a right angle, passing through the y-axis in its way and the resulting vector is $b$ . Then the projection of $3 a + 2 b$ on $c = 5 \hat{i} + 4 \hat{j} + 3 \hat{k}$ is

Select Answer:

Visualized Solution

Visualizing the Initial Setup

Finding the Rotation Plane

Calculating the Plane Vector Components

Evaluating Vector $V$

Determining Vector $b$

Applying the Path Constraint

Setting up the Linear Combination

Calculating the Resultant Vector

Introducing the Target Vector

Computing the Dot Product

Computing the Magnitude of $c$

Final Projection Calculation

The Sigma Insight: Vector Triple Product

Analyzing the Setup

The Vector Triple Product

Normalization and Constraints

Final Calculation

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