Let a be a vector which is perpendicular to the vector 3i + 1/2j + 2k. If a × (2i + k) = 2i - 13j -

Q: Wait, why can't I just use the cross product to find $\vec{a}$ directly? Is it unique?

Great question! The cross product $\vec{a} \times \vec{v} = \vec{w}$ does not define a unique vector $\vec{a}$. It defines a family of vectors that satisfy that specific rotational relationship. The perpendicularity condition $\vec{a} \cdot \vec{v}_1 = 0$ acts as the 'anchor' that selects the one specific vector $\vec{a}$ from that family that satisfies all the problem's requirements.

Q: Why do we get $y=2$ from the cross product? Is that always true?

Yes, it is! When you expand the determinant $\vec{a} \times (2\hat{i} + \hat{k})$, you get expressions for the $\hat{i}, \hat{j},$ and $\hat{k}$ components. By comparing these to the given vector $2\hat{i} - 13\hat{j} - 4\hat{k}$, the $\hat{i}$ component gives $y=2$ and the $\hat{k}$ component gives $-2y = -4$, which also simplifies to $y=2$. This consistency confirms our calculation is on the right track.

Q: Is the projection formula $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$ or $\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|}$?

It is $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$. Think of it geometrically: you are projecting the 'shadow' of vector $\vec{a}$ onto vector $\vec{b}$. Therefore, you must normalize the vector you are projecting onto, which is $\vec{b}$. Always divide by the magnitude of the vector that is receiving the projection.

Question

Let $\vec{a}$ be a vector which is perpendicular to the vector $3\hat{i} + \frac{1}{2}\hat{j} + 2\hat{k}$. If $\vec{a} 	imes (2\hat{i} + \hat{k}) = 2\hat{i} - 13\hat{j} - 4\hat{k}$, then the projection of the vector $\vec{a}$ on the vector $2\hat{i} + 2\hat{j} + \hat{k}$ is

Options:

(A) $\frac{1}{3}$

(B) 1

(C) $\frac{5}{3}$

(D) $\frac{7}{3}$

Sigma Percentile · Accepted Answer

### Defining the Vector

We assume the unknown vector is $\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}$. Our goal is to determine the values of the scalars $x$, $y$, and $z$ using the provided constraints.

### Constraint 1: Perpendicularity

We are given that $\vec{a}$ is perpendicular to $\vec{v}_1 = 3\hat{i} + \frac{1}{2}\hat{j} + 2\hat{k}$. The condition for perpendicularity is that their dot product must be zero:
$$\vec{a} \cdot \vec{v}_1 = 0$$

Expanding this, we have:
$$(x\hat{i} + y\hat{j} + z\hat{k}) \cdot \left(3\hat{i} + \frac{1}{2}\hat{j} + 2\hat{k}ight) = 0$$

This simplifies to $3x + \frac{1}{2}y + 2z = 0$. Multiplying by $2$ to clear the fraction, we obtain **Equation 1**:
$$6x + y + 4z = 0$$

### Constraint 2: The Cross Product

We are given the cross product $\vec{a} 	imes (2\hat{i} + \hat{k}) = 2\hat{i} - 13\hat{j} - 4\hat{k}$. We evaluate the cross product using the determinant method:
$$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ x & y & z \ 2 & 0 & 1 \end{vmatrix} = \hat{i}(y - 0) - \hat{j}(x - 2z) + \hat{k}(0 - 2y)$$

Equating this to the given vector $2\hat{i} - 13\hat{j} - 4\hat{k}$, we compare components:
1. $\hat{i}$ component: $y = 2$
2. $\hat{k}$ component: $-2y = -4 \Rightarrow y = 2$ (Consistent)
3. $\hat{j}$ component: $-(x - 2z) = -13 \Rightarrow x - 2z = 13$ (**Equation 2**)

### Solving the System

Substitute $y = 2$ into **Equation 1**:
$$6x + 2 + 4z = 0 \Rightarrow 6x + 4z = -2$$

Dividing by $2$, we get **Equation 3**:
$$3x + 2z = -1$$

Now, solve the system of **Equation 2** ($x - 2z = 13$) and **Equation 3** ($3x + 2z = -1$):
Adding the two equations:
$$(x - 2z) + (3x + 2z) = 13 - 1$$
$$4x = 12 \Rightarrow x = 3$$

Substitute $x = 3$ into **Equation 2**:
$$3 - 2z = 13 \Rightarrow -2z = 10 \Rightarrow z = -5$$

Thus, the vector is $\vec{a} = 3\hat{i} + 2\hat{j} - 5\hat{k}$.

### Final Calculation: Projection

We find the scalar projection of $\vec{a}$ onto $\vec{b} = 2\hat{i} + 2\hat{j} + \hat{k}$ using the formula:
$$	ext{Projection} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$$

First, calculate the magnitude of $\vec{b}$:
$$|\vec{b}| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4 + 4 + 1} = 3$$

Next, calculate the dot product $\vec{a} \cdot \vec{b}$:
$$\vec{a} \cdot \vec{b} = (3)(2) + (2)(2) + (-5)(1) = 6 + 4 - 5 = 5$$

The final scalar projection is:
$$\mathbf{\frac{5}{3}}$$

Select Answer:

Visualized Solution

Assume $a = x \hat{i} + y \hat{j} + z \hat{k}$

Perpendicularity Condition

Dot Product Equation

Simplifying the Equation

Cross Product Condition

Determinant Expansion

Finding the $y$ Component

Relation between $x$ and $z$

Substituting $y$ in Equation 1

Solving for $x$ and $z$

Defining Vector $a$

Projection Formula

Magnitude of Target Vector

Calculating the Projection

Final Result and Summary

The Sigma Insight: Vector (Cross) Product

Defining the Vector

Constraint 1

Constraint 2

Solving the System

Final Calculation

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