A plane P is parallel to two lines whose direction ratios are -2, 1, -3, and -1, 2, -2 and it contai

Q: Wait, why do we use the cross product here?

Great question! The cross product is the unique tool in vector algebra that generates a vector orthogonal to two given vectors. Since our plane is parallel to the two lines, its normal vector must be orthogonal to the direction vectors of those lines. The cross product is the perfect key to unlock this direction.

Q: Why is the volume formula $\frac{1}{6}|\alpha\beta\gamma|$?

Think of the tetrahedron as a pyramid. The base is a right-angled triangle in the $xy$-plane with area $\frac{1}{2}|\alpha\beta|$, and the height is the intercept on the $z$-axis, $|\gamma|$. The volume of any pyramid is $\frac{1}{3} \times \text{Base Area} \times \text{Height}$. Substituting our values, we get $\frac{1}{3} \times (\frac{1}{2} |\alpha\beta|) \times |\gamma| = \frac{1}{6} |\alpha\beta\gamma|$. It is a beautiful geometric result!

Q: Can I just use the intercept form directly?

Not quite. The intercept form $\frac{x}{\alpha} + \frac{y}{\beta} + \frac{z}{\gamma} = 1$ requires you to know the intercepts first. Since the problem gives you the lines and a point, you must first construct the equation of the plane in the form $Ax + By + Cz = D$ and then manipulate it into the intercept form to extract $\alpha, \beta, \gamma$.

Question

A plane $P$ is parallel to two lines whose direction ratios are $-2, 1, -3$, and $-1, 2, -2$ and it contains the point $(2, 2, -2)$. Let $P$ intersect the co-ordinate axes at the points $A, B, C$ making the intercepts $\alpha, \beta, \gamma$. If $V$ is the volume of the tetrahedron $OABC$, where $O$ is the origin and $p = \alpha + \beta + \gamma$, then the ordered pair $(V, p)$ is equal to

Options:

(A) $(48, -13)$

(B) $(24, -13)$

(C) $(48, 11)$

(D) $(24, -5)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To define a plane in three-dimensional space, we require a point on the plane and a normal vector $\vec{n}$ that dictates its orientation. We are given the point $P_0(2, 2, -2)$.

The plane is parallel to two lines with direction ratios $\vec{d_1} = -2\hat{i} + \hat{j} - 3\hat{k}$ and $\vec{d_2} = -\hat{i} + 2\hat{j} - 2\hat{k}$. Since the plane is parallel to these lines, its normal vector must be perpendicular to both.

### The Normal Vector

We determine the normal vector $\vec{n}$ by calculating the cross product $\vec{d_1} 	imes \vec{d_2}$:

$$
\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -2 & 1 & -3 \ -1 & 2 & -2 \end{vmatrix}
$$

Expanding the determinant, we obtain:

$$
\vec{n} = \hat{i}(-2 - (-6)) - \hat{j}(4 - 3) + \hat{k}(-4 - (-1))
$$

This simplifies to the normal vector $\vec{n} = 4\hat{i} - \hat{j} - 3\hat{k}$. This vector serves as the fundamental orientation of our plane.

### Building the Plane Equation

Using the point-normal form $a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$ with point $(2, 2, -2)$ and normal $(4, -1, -3)$, we write:

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

Expanding the terms yields $4x - 8 - y + 2 - 3z - 6 = 0$. Simplifying this, we arrive at the standard form of the plane:

$$
4x - y - 3z = 12
$$

### Intercepts and Final Calculation

To find the intercepts, we convert the equation to the intercept form $\frac{x}{\alpha} + \frac{y}{\beta} + \frac{z}{\gamma} = 1$ by dividing by $12$:

$$
\frac{x}{3} + \frac{y}{-12} + \frac{z}{-4} = 1
$$

From this, the intercepts are $\alpha = 3$, $\beta = -12$, and $\gamma = -4$. The sum of these intercepts is $p = 3 - 12 - 4 = -13$.

The volume $V$ of the tetrahedron formed by the origin and these intercepts is:

$$
V = \frac{1}{6} |\alpha \beta \gamma| = \frac{1}{6} |3 	imes (-12) 	imes (-4)| = \frac{144}{6} = 24
$$

The final result, expressed as the ordered pair $(V, p)$, is **$(24, -13)$**.

Select Answer:

Visualized Solution

The 3D Setup

The Normal Vector Concept

Setting up the Cross Product

Calculating Final Normal Vector

Point-Normal Form of Plane

Substituting Values into Plane Equation

Expanding the Equation

Standard Form of Plane

Intercept Form

Finding Intercepts

Extracting $α, β, γ$

Calculating $p$

Volume of Tetrahedron

Calculating Volume $V$

Final Answer

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Normal Vector

Building the Plane Equation

Intercepts and Final Calculation

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