Let O be the origin. Let OP = xi + yj - k and OQ = -i + 2j + 3xk, x, y ∈ R, x 0, be such that |PQ|

Q: Wait, why do we use the determinant for coplanarity? Can't we just use the cross product?

That is a fantastic observation! You absolutely can use the cross product. If $\vec{OP}$ and $\vec{OQ}$ are perpendicular, their cross product $\vec{OP} \times \vec{OQ}$ gives a vector normal to the plane containing them. For $\vec{OR}$ to be coplanar, it must be perpendicular to that normal vector. So, $(\vec{OP} \times \vec{OQ}) \cdot \vec{OR} = 0$. The determinant is just a compact, elegant way of writing this exact scalar triple product. It saves time and reduces the chance of algebraic errors!

Q: Why did we reject the negative root for $x$? The equation $x^2 = 1$ gives $x = 1$ or $x = -1$.

Great catch! In the heat of solving, it is easy to forget the constraints given in the problem statement. The question explicitly states $x > 0$. In JEE Advanced, these constraints are not just filler text; they are boundary conditions that define the physical reality of the problem. Always check your constraints before moving to the next step!

Q: I keep getting a different value for $z$. Is there a trick to the determinant expansion?

Determinant expansion is a classic place for 'sign traps.' When you expand, remember the alternating signs: $+ - +$. For the second term in the first row, you must multiply by $-1$. If you missed that negative sign, your equation for $z$ will be incorrect. Take a deep breath, re-expand slowly, and watch those signs like a hawk!

Question

Let O be the origin. Let $\vec{OP} = x\hat{i} + y\hat{j} - \hat{k}$ and $\vec{OQ} = -\hat{i} + 2\hat{j} + 3x\hat{k}$, $x, y \in R, x > 0$, be such that $|\vec{PQ}| = \sqrt{20}$ and the vector $\vec{OP}$ is perpendicular to $\vec{OQ}$. If $\vec{OR} = 3\hat{i} + z\hat{j} - 7\hat{k}$, $z \in R$, is coplanar with $\vec{OP}$ and $\vec{OQ}$, then the value of $x^2 + y^2 + z^2$ is equal to

Options:

(A) 7

(B) 9

(C) 2

(D) 1

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not just solving a problem; we are navigating the architecture of three-dimensional space. When you look at a problem involving vectors like $\vec{OP}$, $\vec{OQ}$, and $\vec{OR}$, do not see them as mere collections of numbers.

See them as arrows piercing through the fabric of space, originating from the origin $O$. Our mission is to uncover the hidden values of $x$, $y$, and $z$ that define these vectors, and ultimately, to find the sum of their squares. Let us embark on this journey step by step.

### The Perpendicularity Dance

The problem begins with a powerful constraint: $\vec{OP}$ is perpendicular to $\vec{OQ}$. In the language of physics and mathematics, perpendicularity is synonymous with the dot product being zero.

Because the dot product is defined as $\vec{A} \cdot \vec{B} = |\vec{A}||\vec{B}| \cos(	heta)$, if the angle $	heta$ is $90^{\circ}$, then $\cos(90^{\circ}) = 0$, and the entire product vanishes. We are given:

$$\vec{OP} = x\hat{i} + y\hat{j} - \hat{k}$$
$$\vec{OQ} = -\hat{i} + 2\hat{j} + 3x\hat{k}$$

Applying the dot product condition $\vec{OP} \cdot \vec{OQ} = 0$, we multiply the corresponding components:

$$(x)(-1) + (y)(2) + (-1)(3x) = 0$$

This simplifies to $-x + 2y - 3x = 0$, which further reduces to $2y - 4x = 0$. With a simple algebraic step, we find the beautiful relationship $y = 2x$. This is our first major breakthrough, as we have reduced the number of unknowns by linking $y$ directly to $x$.

### The Magnitude Constraint

Now, we turn our attention to the magnitude of the vector $\vec{PQ}$. The problem tells us that $|\vec{PQ}| = \sqrt{20}$. First, we must construct the vector $\vec{PQ}$ using the triangle law of vector addition: $\vec{PQ} = \vec{OQ} - \vec{OP}$.

Subtracting the components, we get:

$$\vec{PQ} = (-1 - x)\hat{i} + (2 - y)\hat{j} + (3x + 1)\hat{k}$$

To find the magnitude, we square the components and sum them. Since $|\vec{PQ}| = \sqrt{20}$, it follows that $|\vec{PQ}|^2 = 20$:

$$(-1 - x)^2 + (2 - y)^2 + (3x + 1)^2 = 20$$

Here is where our earlier discovery, $y = 2x$, becomes our best friend. We substitute $y$ with $2x$ to create an equation entirely in terms of $x$:

$$(x + 1)^2 + (2 - 2x)^2 + (3x + 1)^2 = 20$$

Expanding these binomials requires patience:

$$(x^2 + 2x + 1) + (4 - 8x + 4x^2) + (9x^2 + 6x + 1) = 20$$

Grouping the terms, we get $14x^2 + 0x + 6 = 20$. This simplifies to $14x^2 = 14$, or $x^2 = 1$.

Since the problem states $x > 0$, we must reject $x = -1$ and accept $x = 1$. With $x = 1$, we immediately find $y = 2(1) = 2$. We have conquered the first two variables!

### The Coplanarity Mystery

Finally, we address the condition that $\vec{OR} = 3\hat{i} + z\hat{j} - 7\hat{k}$ is coplanar with $\vec{OP}$ and $\vec{OQ}$. When three vectors are coplanar, they lie in the same flat plane. Mathematically, this means the volume of the parallelepiped formed by them is zero, expressed by the scalar triple product being zero:

$$[\vec{OP} \ \vec{OQ} \ \vec{OR}] = 0$$

We set up the determinant using the components of our vectors:

$$\begin{vmatrix} 1 & 2 & -1 \ -1 & 2 & 3 \ 3 & z & -7 \end{vmatrix} = 0$$

Expanding along the first row:

$$1((2)(-7) - (3)(z)) - 2((-1)(-7) - (3)(3)) - 1((-1)(z) - (2)(3)) = 0$$

Let us simplify this step-by-step:

$$1(-14 - 3z) - 2(7 - 9) - 1(-z - 6) = 0$$
$$-14 - 3z + 4 + z + 6 = 0$$
$$-4 - 2z = 0$$

Solving for $z$, we get $-2z = 4$, which means $z = -2$.

### Final Calculation

We have arrived at the finish line. We have found $x = 1$, $y = 2$, and $z = -2$. The question asks for the value of $x^2 + y^2 + z^2$.

Substituting our values:

$$x^2 + y^2 + z^2 = (1)^2 + (2)^2 + (-2)^2$$
$$= 1 + 4 + 4 = 9$$

And there it is—the value is **9**. You have navigated the perpendicularity, the magnitude, and the coplanarity conditions with precision. Remember, in JEE Advanced, the math is just the language; the real skill is in the logical flow and the careful execution.

Select Answer:

Visualized Solution

Visualizing the Vectors

Applying Perpendicularity

Dot Product Calculation

Finding Vector $P Q$

Magnitude of $P Q$

Substitution and Expansion

Solving for $x$

Finding $y$ and Updating Vectors

Condition of Coplanarity

Expanding the Determinant

Solving for $z$

Final Calculation

The Sigma Insight: Scalar Triple Product

Analyzing the Setup

The Perpendicularity Dance

The Magnitude Constraint

The Coplanarity Mystery

Final Calculation

Similar Questions

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Select Answer:

Visualized Solution

Visualizing the Vectors

Applying Perpendicularity

Dot Product Calculation

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

Magnitude of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PQ​

Substitution and Expansion

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

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

Condition of Coplanarity

Expanding the Determinant

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

Final Calculation

The Sigma Insight: Scalar Triple Product

Analyzing the Setup

The Perpendicularity Dance

The Magnitude Constraint

The Coplanarity Mystery

Final Calculation

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