Let ABC be a triangle of area 15√2 and the vectors AB = i + 2j - 7k, BC = ai + bj + ck and AC = 6i +

Q: Wait, why do we use the cross product for area? Can't we use the dot product?

Great question! The dot product gives us the cosine of the angle between vectors, which is useful for projections or work done. But for area, we need the sine of the angle, which is exactly what the magnitude of the cross product provides. It is the geometric definition of the area of a parallelogram!

Q: Why do we reject the negative root for $d$?

The problem statement is our law. It explicitly mentions $d > 0$. In JEE problems, constraints are not just suggestions; they are filters. Always check your final values against the given constraints to ensure your solution is physically and mathematically valid.

Q: Is there a faster way to find the third side $\vec{BC}$?

Yes! The triangle law of vector addition is your best friend. Since $\vec{AB} + \vec{BC} = \vec{AC}$, it follows that $\vec{BC} = \vec{AC} - \vec{AB}$. This is much faster than calculating coordinates or using distance formulas from scratch.

Question

Let $ABC$ be a triangle of area $15\sqrt{2}$ and the vectors $\vec{AB} = \hat{i} + 2\hat{j} - 7\hat{k}$, $\vec{BC} = a\hat{i} + b\hat{j} + c\hat{k}$ and $\vec{AC} = 6\hat{i} + d\hat{j} - 2\hat{k}$, $d > 0$. Then the square of the length of the largest side of the triangle $ABC$ is _______

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given a triangle $ABC$ in 3D space with vectors $\vec{AB} = \hat{i} + 2\hat{j} - 7\hat{k}$ and $\vec{AC} = 6\hat{i} + d\hat{j} - 2\hat{k}$. The area of this triangle is provided as $15\sqrt{2}$.

The area of a triangle defined by two vectors is given by the formula:
$$	ext{Area} = \frac{1}{2} |\vec{AB} 	imes \vec{AC}|$$

This relationship exists because the magnitude of the cross product represents the area of the parallelogram formed by the two vectors, and the triangle occupies exactly half of that space.

### The Master Equation

To find the cross product, we evaluate the following determinant:
$$\vec{AB} 	imes \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & -7 \ 6 & d & -2 \end{vmatrix}$$

Expanding this determinant along the first row, we obtain:
$$\vec{AB} 	imes \vec{AC} = ((-4) - (-7d))\hat{i} - ((-2) - (-42))\hat{j} + (d - 12)\hat{k}$$
$$\vec{AB} 	imes \vec{AC} = (7d - 4)\hat{i} - 40\hat{j} + (d - 12)\hat{k}$$

### Solving for the Variable

Given the area is $15\sqrt{2}$, we set the magnitude of the cross product equal to $30\sqrt{2}$:
$$|(7d - 4)\hat{i} - 40\hat{j} + (d - 12)\hat{k}| = 30\sqrt{2}$$

Squaring both sides to eliminate the square root, we get:
$$(7d - 4)^2 + (-40)^2 + (d - 12)^2 = (30\sqrt{2})^2$$
$$(49d^2 - 56d + 16) + 1600 + (d^2 - 24d + 144) = 1800$$

Simplifying the quadratic equation:
$$50d^2 - 80d - 40 = 0$$
$$5d^2 - 8d - 4 = 0$$

Factoring the quadratic equation $(5d + 2)(d - 2) = 0$, we find the roots $d = 2$ and $d = -2/5$. Applying the constraint $d > 0$, we conclude that **$d = 2$**.

### Final Calculation

With $d = 2$, we determine the vector $\vec{BC}$ using the triangle law of vector addition:
$$\vec{BC} = \vec{AC} - \vec{AB} = (6-1)\hat{i} + (2-2)\hat{j} + (-2 - (-7))\hat{k}$$
$$\vec{BC} = 5\hat{i} + 0\hat{j} + 5\hat{k}$$

Finally, we calculate the squared lengths of the sides:
$$|\vec{AB}|^2 = 1^2 + 2^2 + (-7)^2 = 1 + 4 + 49 = 54$$
$$|\vec{AC}|^2 = 6^2 + 2^2 + (-2)^2 = 36 + 4 + 4 = 44$$
$$|\vec{BC}|^2 = 5^2 + 0^2 + 5^2 = 25 + 0 + 25 = 50$$

The largest squared length is **54**.

Enter Numerical Value:

Visualized Solution

Visualizing Triangle $A B C$

Area Formula using Cross Product

Setting up the Cross Product

Expanding the Determinant

Relating Magnitude to Area

Squaring the Magnitude Equation

Simplifying the Quadratic Equation

Solving for $d$

Finding the Third Side $B C$

Calculating Squared Length of $A B$

Calculating Squared Length of $A C$

Calculating Squared Length of $B C$

The Final Answer

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Master Equation

Solving for the Variable

Final Calculation

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