If the volume of a parallelopiped, whose coterminus edges are given by the vectors a = i + j + nk, b

Q: Wait, why do we use the determinant for volume? Can't we just use base area times height?

You absolutely can! The scalar triple product $\vec{a} \cdot (\vec{b} \times \vec{c})$ is mathematically equivalent to the base area (the magnitude of the cross product $\vec{b} \times \vec{c}$) multiplied by the height (the projection of $\vec{a}$ onto the normal of the base). The determinant is simply the most efficient computational tool to calculate this product directly without finding the cross product first.

Q: Why did we reject the negative root for $n$? Is it possible for a vector component to be negative?

A vector component can absolutely be negative! However, the problem statement explicitly specifies the constraint $n > 0$. In JEE problems, always keep an eye on these constraints—they are often the key to selecting the correct root in quadratic equations.

Q: Is there a faster way to check the options instead of solving the whole quadratic?

You could potentially test the options for $n$ by plugging them into the determinant, but that would require calculating a $3 \times 3$ determinant for each option. Solving the quadratic equation is actually the most robust and reliable method here. Once you have $n=8$, verifying the dot product is trivial.

Question

If the volume of a parallelopiped, whose coterminus edges are given by the vectors $\vec{a} = \hat{i} + \hat{j} + n\hat{k}, \vec{b} = 2\hat{i} + 4\hat{j} - n\hat{k}$ and $\vec{c} = \hat{i} + n\hat{j} + 3\hat{k} (n > 0)$, is 158 cu.units, then :

Options:

(A) $n = 7$

(B) $\vec{b} \cdot \vec{c} = 10$

(C) $n = 9$

(D) $\vec{a} \cdot \vec{c} = 17$

Sigma Percentile · Accepted Answer

### The Geometry of Space: Unlocking the Parallelepiped

Welcome, future engineer. Today, we are not just solving a vector problem; we are stepping into the realm of three-dimensional geometry. Imagine you are standing in a room, and in the corner, you see three vectors, $\vec{a}, \vec{b},$ and $\vec{c}$, originating from a single point.

These are your coterminous edges. When you connect them, they define a slanted, skewed box—a parallelepiped. This shape is the fundamental building block of 3D vector analysis, and understanding its volume is a rite of passage for every JEE aspirant.

### The Scalar Triple Product: The Engine of Volume

We are given the volume of this parallelepiped as $158$ cubic units. The volume $V$ of a parallelepiped defined by vectors $\vec{a}, \vec{b},$ and $\vec{c}$ is given by the absolute value of the scalar triple product: $V = |[\vec{a} \vec{b} \vec{c}]|$.

Mathematically, this product is the determinant of the matrix formed by the components of these vectors. We set up our matrix using the given vectors:

$$\begin{vmatrix} 1 & 1 & n \ 2 & 4 & -n \ 1 & n & 3 \end{vmatrix} = 158$$

I know that looking at a determinant with a variable $n$ can feel intimidating. But take a deep breath; this is just a systematic process of expansion. We are going to break this down, step by step, and watch the complexity collapse into a simple quadratic equation.

### The Algebraic Grind: Expanding the Determinant

Let us expand this determinant along the first row. We take the first element, $1$, and multiply it by the determinant of the minor matrix, then subtract the second element, $1$, multiplied by its minor, and finally add the third element, $n$, multiplied by its minor.

$$1 \cdot (12 - (-n^2)) - 1 \cdot (6 - (-n)) + n \cdot (2n - 4) = 158$$

Simplifying the terms inside the parentheses gives us:

$$1(12 + n^2) - 1(6 + n) + n(2n - 4) = 158$$

Now, let us distribute the coefficients to organize the expression:

$$12 + n^2 - 6 - n + 2n^2 - 4n = 158$$

Grouping the like terms is the next logical step. We combine the $n^2$ terms, the $n$ terms, and the constants:

$$(n^2 + 2n^2) + (-n - 4n) + (12 - 6) = 158$$

$$3n^2 - 5n + 6 = 158$$

Subtracting $158$ from both sides brings us to our final quadratic equation:

$$3n^2 - 5n - 152 = 0$$

### The Resolution: Solving for $n$

We have arrived at a quadratic equation. To solve for $n$, we use the quadratic formula: $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Here, $a = 3$, $b = -5$, and $c = -152$.

Substituting these values in:

$$n = \frac{5 \pm \sqrt{(-5)^2 - 4(3)(-152)}}{2(3)}$$

$$n = \frac{5 \pm \sqrt{25 + 1824}}{6}$$

$$n = \frac{5 \pm \sqrt{1849}}{6}$$

Calculating the square root of $1849$ yields $43$. Thus, we have two potential solutions:

$$n = \frac{5 + 43}{6} = \frac{48}{6} = 8$$

$$n = \frac{5 - 43}{6} = -\frac{38}{6} = -\frac{19}{3}$$

Recall the constraint given in the problem: $n > 0$. This is the crucial filter. We must reject the negative root. Therefore, our valid value is **$n = 8$**.

### Verification: The Final Check

With $n = 8$, we can now define our vectors completely:

$$\vec{b} = 2\hat{i} + 4\hat{j} - 8\hat{k}$$

$$\vec{c} = \hat{i} + 8\hat{j} + 3\hat{k}$$

Let us verify the condition $\vec{b} \cdot \vec{c} = 10$. The dot product is calculated by summing the products of the corresponding components:

$$\vec{b} \cdot \vec{c} = (2)(1) + (4)(8) + (-8)(3)$$

$$\vec{b} \cdot \vec{c} = 2 + 32 - 24 = 10$$

It matches perfectly! We have navigated the geometry, conquered the algebra, and verified our result. Remember, in JEE Advanced, the math is just the language; the true skill is in the visualization and the careful execution of the steps.

Animated Solution for Mathematics - Vector Algebra: If the volume of a parallelopiped, whose coterminus edges are given by the vectors $a = \hat{i} + \hat{j} + n \hat{k}, b = 2 \hat{i} + 4 \hat{j} - n \hat{k}$ and $c = \hat{i} + n \hat{j} + 3 \hat{k} (n > 0)$ , is 158 cu.units, then :

Select Answer:

Visualized Solution

Visualizing the Parallelepiped

The Scalar Triple Product Formula

Expanding the Determinant

Simplifying to a Quadratic Equation

Solving the Quadratic Equation

Finding the Valid Value of $n$

Checking the Options

Final Conclusion

The Sigma Insight: Scalar Triple Product

The Geometry of Space

The Scalar Triple Product

The Algebraic Grind

The Resolution

Verification

Similar Questions

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If $a = 2 \hat{i} + \hat{j} + 3 \hat{k}, b = 3 \hat{i} + 3 \hat{j} + \hat{k}$ and $c = c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k}$ are coplanar vectors and $a \cdot c = 5, b ⊥ c$ , then $122 (c_{1} + c_{2} + c_{3})$ is equal to

Match List I with List II:

List-I

List-II