Let a = i + 2j + k and b = 2i + j - k. Let c be a unit vector in the plane of the vectors a and b an

Q: Wait, why can't I just use $\vec{a} \times \vec{b}$? Isn't that perpendicular to the plane?

That is a fantastic question! You are absolutely right that $\vec{a} \times \vec{b}$ is perpendicular to the plane containing $\vec{a}$ and $\vec{b}$. However, the problem asks for a vector that lies *in* the plane. If you use the cross product of $\vec{a}$ and $\vec{b}$, you are pointing out of the plane, not along it. By taking the cross product of that normal vector with $\vec{a}$ again, we rotate our direction back into the plane.

Q: Why do we use the BAC-CAB rule instead of just calculating the cross product directly?

You certainly *could* calculate the cross product directly using determinants, but it is prone to arithmetic errors. The BAC-CAB rule, $(\vec{a} \times \vec{b}) \times \vec{a} = (\vec{a} \cdot \vec{a})\vec{b} - (\vec{a} \cdot \vec{b})\vec{a}$, turns a complex vector operation into simple scalar multiplication and vector subtraction. It is a much more elegant and efficient path to the solution.

Q: I got $\frac{1}{\sqrt{2}}(\hat{i} - \hat{k})$, but the answer is $\frac{1}{\sqrt{2}}(-\hat{i} + \hat{k})$. Did I do something wrong?

Not at all! Remember that a unit vector can point in either of two opposite directions along a line. If your vector is $\vec{v}$, then both $\frac{\vec{v}}{|\vec{v}|}$ and $-\frac{\vec{v}}{|\vec{v}|}$ are valid unit vectors. Since the question asks for 'such a vector,' it implies that one of the two possible directions will be listed in the options. You simply found the opposite direction, which is perfectly valid.

Question

Let $\vec{a} = \hat{i} + 2\hat{j} + \hat{k}$ and $\vec{b} = 2\hat{i} + \hat{j} - \hat{k}$. Let $\hat{c}$ be a unit vector in the plane of the vectors $\vec{a}$ and $\vec{b}$ and be perpendicular to $\vec{a}$. Then such a vector $\hat{c}$ is :

Options:

(A) $\frac{1}{\sqrt{5}}(\hat{j} - 2\hat{k})$

(B) $\frac{1}{\sqrt{3}}(-\hat{i} + \hat{j} - \hat{k})$

(C) $\frac{1}{\sqrt{3}}(\hat{i} - \hat{j} + \hat{k})$

(D) $\frac{1}{\sqrt{2}}(-\hat{i} + \hat{k})$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional room. You have two vectors, $\vec{a} = \hat{i} + 2\hat{j} + \hat{k}$ and $\vec{b} = 2\hat{i} + \hat{j} - \hat{k}$, originating from the same point.

These two vectors define a flat, two-dimensional surface—a plane—stretching out into the room. Our mission is to find a unit vector $\hat{c}$ that lies on this plane and is perfectly perpendicular to $\vec{a}$.

### The Magic of the Vector Triple Product

How do we mathematically trap a vector on a plane while forcing it to be perpendicular to another? We need a tool that respects the geometry of the plane.

The vector triple product, $(\vec{a} 	imes \vec{b}) 	imes \vec{a}$, is our secret weapon. The term $\vec{a} 	imes \vec{b}$ creates a normal vector, sticking straight out of the plane.

When we cross that normal vector with $\vec{a}$ again, we rotate our direction by $90^{\circ}$, forcing the result to lie back within the plane of $\vec{a}$ and $\vec{b}$. It is a beautiful, self-correcting mechanism.

### Simplifying with BAC-CAB

We can use the elegant **BAC-CAB identity** to simplify the expression:

$$(\vec{a} 	imes \vec{b}) 	imes \vec{a} = (\vec{a} \cdot \vec{a})\vec{b} - (\vec{a} \cdot \vec{b})\vec{a}$$

First, we calculate the dot product $\vec{a} \cdot \vec{a}$, which is the square of the magnitude of $\vec{a}$. With $\vec{a} = \hat{i} + 2\hat{j} + \hat{k}$, we find:

$$\vec{a} \cdot \vec{a} = (1)^2 + (2)^2 + (1)^2 = 6$$

Next, we calculate the dot product $\vec{a} \cdot \vec{b}$:

$$\vec{a} \cdot \vec{b} = (1)(2) + (2)(1) + (1)(-1) = 2 + 2 - 1 = 3$$

### The Final Calculation

Substituting these values back into our expansion, we define the vector $\vec{v}$ as:

$$\vec{v} = 6\vec{b} - 3\vec{a}$$

Plugging in the components, we have:

$$\vec{v} = 6(2\hat{i} + \hat{j} - \hat{k}) - 3(\hat{i} + 2\hat{j} + \hat{k})$$

Distributing the scalars and subtracting component-wise, the $\hat{j}$ terms cancel out:

$$\vec{v} = (12 - 3)\hat{i} + (6 - 6)\hat{j} + (-6 - 3)\hat{k} = 9\hat{i} - 9\hat{k}$$

We have found a vector $\vec{v} = 9(\hat{i} - \hat{k})$ that is parallel to our desired unit vector $\hat{c}$.

### Normalizing to Unity

The final step is to turn this into a unit vector by dividing $\vec{v}$ by its magnitude. The magnitude of $\hat{i} - \hat{k}$ is:

$$|\hat{i} - \hat{k}| = \sqrt{1^2 + (-1)^2} = \sqrt{2}$$

Thus, the unit vector is:

$$\hat{c} = \pm \frac{1}{\sqrt{2}}(\hat{i} - \hat{k})$$

The result $\frac{1}{\sqrt{2}}(-\hat{i} + \hat{k})$ is simply the negative version of our result, which is perfectly valid. We have successfully navigated the geometry and arrived at the **correct orientation**.

Animated Solution for Mathematics - Vector Algebra: Let $a = \hat{i} + 2 \hat{j} + \hat{k}$ and $b = 2 \hat{i} + \hat{j} - \hat{k}$ . Let $\overset{c}{^}$ be a unit vector in the plane of the vectors $a$ and $b$ and be perpendicular to $a$ . Then such a vector $\overset{c}{^}$ is :

Select Answer:

Visualized Solution

Visualizing the Given Vectors

The Constraints on Vector $\overset{c}{^}$

The Vector Triple Product

Expanding the Triple Product

Calculating $a \cdot a$

Calculating $a \cdot b$

Substituting into the Formula

Expanding the Expression

Simplifying the Vector

Finding the Unit Vector

Final Answer Selection

The Sigma Insight: Vector Triple Product

Analyzing the Setup

The Magic of the Vector Triple Product

Simplifying with BAC-CAB

The Final Calculation

Normalizing to Unity

Similar Questions

Let $x, y$ and $z$ be three vectors each of magnitude $2$ and the angle between each pair of them is $\frac{π}{3}$ . If $a$ is a non-zero vector perpendicular to $x$ and $y \times z$ and $b$ is a non-zero vector perpendicular to $y$ and $z \times x$ , then

A unit vector coplanar with $\hat{i} + \hat{j} + 2 \hat{k}$ and $\hat{i} + 2 \hat{j} + \hat{k}$ and perpendicular to $\hat{i} + \hat{j} + \hat{k}$ is .........

Let $a = \hat{i} - 2 \hat{j} + \hat{k}$ and $b = \hat{i} - \hat{j} + \hat{k}$ be two vectors. If $c$ is a vector such that $b \times c = b \times a$ and $c \cdot a = 0$ , then $c \cdot b$ is equal to

Let $a = 2 \hat{i} + \hat{j} - 2 \hat{k}$ and $b = \hat{i} + \hat{j}$ . Let $c$ be a vector such that $∣ c - a ∣ = 3$ , $∣ (a \times b) \times c ∣ = 3$ and the angle between $c$ and $a \times b$ be $3 0^{\circ}$ . Then $a \cdot c$ is equal to:

Let $a = 2 \hat{i} - \hat{j} + 2 \hat{k}$ and $b = \hat{i} + 2 \hat{j} - \hat{k}$ . Let a vector $v$ be in the plane containing $a$ and $b$ . If $v$ is perpendicular to the vector $3 \hat{i} + 2 \hat{j} - \hat{k}$ and its projection on $a$ is 19 units, then $∣2 v ∣^{2}$ is equal to .

Let $a, b, c$ be three vectors mutually perpendicular to each other and have same magnitude. If a vector $r$ satisfies $a \times {(r - b) \times a} + b \times {(r - c) \times b} + c \times {(r - a) \times c} = 0$ , then $r$ is equal to :

If $a = (\hat{i} + \hat{j} + \hat{k})$ , $a \cdot b = 1$ and $a \times b = \hat{j} - \hat{k}$ , then $b$ is

Let $a = - 5 \hat{i} + \hat{j} - 3 \hat{k}$ , $b = \hat{i} + 2 \hat{j} - 4 \hat{k}$ and $c = (((a \times b) \times \hat{i}) \times \hat{i}) \times \hat{i}$ . Then $c \cdot (- \hat{i} + \hat{j} + \hat{k})$ is equal to

If $a, b, c$ are three non-zero vectors and $\overset{n}{^}$ is a unit vector perpendicular to $c$ such that $a = α b - \overset{n}{^}, (α \neq = 0)$ and $b \cdot c = 12$ , then $∣ c \times (a \times b) ∣$ is equal to:

Let $a, b$ and $c$ be three unit vectors such that $a \times (b \times c) = \frac{3}{2} (b + c)$ . If $b$ is not parallel to $c$ , then the angle between $a$ and $b$ is: