A unit vector coplanar with i + j + 2k and i + 2j + k and perpendicular to i + j + k is .........

Q: Wait, why is the scalar triple product zero for coplanar vectors? I thought that was only for parallel vectors?

That is a fantastic question! Think of the Scalar Triple Product $[\vec{r} \quad \vec{a} \quad \vec{b}]$ as the volume of a parallelepiped defined by these three vectors. If the vectors are coplanar, they all lie on the same flat sheet. A flat sheet has no 'height' or 'thickness' in the third dimension, so the volume of the shape they form collapses to zero. It is a beautiful geometric proof of coplanarity!

Q: Why are there two possible answers for the vector? Did I miss a step?

You didn't miss a step at all—you actually found the full solution! When we solve for the magnitude constraint $|\vec{r}| = 1$, we end up with $\lambda^2 = \frac{1}{2}$, which gives $\lambda = \pm \frac{1}{\sqrt{2}}$. This means there are two unit vectors that satisfy the conditions: one pointing in one direction along the line of intersection, and one pointing in the exact opposite direction. Both are perfectly valid!

Q: Can I use the cross product instead of the determinant to find the coplanar vector?

Absolutely! If $\vec{r}$ is coplanar with $\vec{a}$ and $\vec{b}$, then $\vec{r}$ must be parallel to the vector $\vec{a} \times \vec{b}$. You could calculate $\vec{n} = \vec{a} \times \vec{b}$ and then find a vector perpendicular to both $\vec{n}$ and $\vec{c}$. Both methods are mathematically sound and lead to the same result. The determinant method is just a more direct way to handle the coplanarity condition.

Question

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 .........

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers. Today, we are not just solving a math problem; we are navigating the architecture of 3D space. When you look at a problem involving vectors, I want you to stop seeing just numbers and variables.

I want you to see arrows in space, planes slicing through the coordinate system, and the elegant constraints that bind them together. We are looking for a mysterious unit vector, $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$, that satisfies two very specific geometric conditions.

### The Coplanarity Trap

The first condition is that our vector $\vec{r}$ is coplanar with $\vec{a} = \hat{i} + \hat{j} + 2\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$. If $\vec{a}$ and $\vec{b}$ are drawn on a sheet of paper, and $\vec{r}$ is also on that paper, they are coplanar.

In the language of JEE Advanced, we translate this geometric reality into the Scalar Triple Product. If three vectors are coplanar, they cannot form a 3D volume, meaning the parallelepiped they define has a volume of zero:

$$[\vec{r} \quad \vec{a} \quad \vec{b}] = 0$$

We set up our determinant, placing the components of $\vec{r}$ in the first row, and the components of $\vec{a}$ and $\vec{b}$ in the subsequent rows:

$$\begin{vmatrix} x & y & z \ 1 & 1 & 2 \ 1 & 2 & 1 \end{vmatrix} = 0$$

Expanding this along the first row, we calculate the minors: $x(1-4) - y(1-2) + z(2-1) = 0$. Simplifying this, we arrive at our first linear equation:

$$-3x + y + z = 0$$

### The Perpendicularity Bridge

Now, we introduce the second condition. Our vector $\vec{r}$ is perpendicular to $\vec{c} = \hat{i} + \hat{j} + \hat{k}$. In the world of vectors, perpendicularity is synonymous with the dot product being zero.

So, we write:

$$\vec{r} \cdot \vec{c} = 0$$

Substituting our components, we get $(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k}) = 0$, which simplifies beautifully to:

$$x + y + z = 0$$

### Solving the System

We have three variables but only two equations. This means our vector $\vec{r}$ is not unique in terms of magnitude, but it is unique in terms of **direction**. We can find the ratio of $x:y:z$ using the method of cross-multiplication.

Setting up the ratios:

$$\frac{x}{(1)(1) - (1)(1)} = \frac{-y}{(-3)(1) - (1)(1)} = \frac{z}{(-3)(1) - (1)(1)}$$

This simplifies to $\frac{x}{0} = \frac{-y}{-4} = \frac{z}{-4}$. Dividing by $4$, we get the direction ratios: $\frac{x}{0} = \frac{y}{1} = \frac{z}{-1}$. We can now define our vector in terms of a scalar parameter $\lambda$:

$$\vec{r} = \lambda(0\hat{i} + 1\hat{j} - 1\hat{k})$$

### The Final Constraint

The problem specifically asked for a **unit vector**. This is our final gatekeeper, requiring the magnitude of $\vec{r}$ to be $1$:

$$\sqrt{x^2 + y^2 + z^2} = 1$$

Substituting our values in terms of $\lambda$ ($x=0, y=\lambda, z=-\lambda$):

$$\sqrt{0^2 + \lambda^2 + (-\lambda)^2} = 1 \Rightarrow \sqrt{2\lambda^2} = 1 \Rightarrow 2\lambda^2 = 1$$

This gives us $\lambda^2 = \frac{1}{2}$, so $\lambda = \pm \frac{1}{\sqrt{2}}$. By substituting these values back into our expression for $\vec{r}$, we find our two solutions:

$$\vec{r} = \pm \frac{\hat{j} - \hat{k}}{\sqrt{2}}$$

### Conclusion

You took an abstract geometric description, translated it into the language of determinants and dot products, solved a system of equations, and applied a normalization constraint. This is the essence of JEE Advanced mathematics—taking a complex, multi-layered problem and peeling it back until the truth is revealed. Keep practicing this systematic approach, and you will find that no problem is too daunting.

Animated Solution for Mathematics - Vector Algebra: 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 .........

Visualized Solution

Defining the Unknown Vector $r$

The Condition of Coplanarity

Setting up the Determinant

Expanding the Determinant

The Condition of Perpendicularity

Setting up the Dot Product

Solving the System of Equations

Cross-Multiplication Method

Finding the Direction Ratios

Applying the Unit Vector Constraint

Solving for $λ$

Final Result

The Sigma Insight: Vector Triple Product

Analyzing the Setup

The Coplanarity Trap

The Perpendicularity Bridge

Solving the System

The Final Constraint

Conclusion

Similar Questions

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 :

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

Let three vectors $a, b$ and $c$ be such that $c$ is coplanar with $a$ and $b, a \cdot c = 7$ and $b$ is perpendicular to $c$ , where $a = - \hat{i} + \hat{j} + \hat{k}$ and $b = 2 \hat{i} + \hat{k}$ , then the value of $2∣ a + b + c ∣^{2}$ is

If $a, b, c$ are non coplanar unit vectors such that $a \times (b \times c) = \frac{( b + c )}{2}$ , then the angle between $a$ and $b$ is

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 .

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 $a, b, c$ 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 :

Let $u$ and $v$ be unit vectors. If $w$ is a vector such that $w + (w \times u) = v$ , then prove that $∣ (u \times v) \cdot w ∣ \leq 1/2$ and that the equality holds if and only if $u$ is perpendicular to $v$ .

If the vectors $b, c, d$ are not coplanar, then prove that the vector $(a \times b) \times (c \times d) + (a \times c) \times (d \times b) + (a \times d) \times (b \times c)$ is parallel to $a$ .

If $a$ and $b$ are perpendicular vectors, then $a \times (a \times (a \times (a \times b)))$ is equal to

Animated Solution for Mathematics - Vector Algebra: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A unit vector coplanar with i^+j^​+2k^ and i^+2j^​+k^ and perpendicular to i^+j^​+k^ is .........