Let a line pass through two distinct points P(-2,-1,3) and Q, and be parallel to the vector 3i+2j+2k

Q: Why do we reject \lambda = 0?

The problem explicitly states that P and Q are distinct points. If \lambda = 0, then Q = P, which would make the triangle degenerate and violate the condition of distinct points. Always check your constraints against the physical reality of the problem!

Q: Can I use the formula for the distance from a point to a line?

You could, but that formula gives you the *minimum* distance from a point to a line. Here, we are looking for a specific point Q at a fixed distance of 5, so the parametric approach is much more direct and less prone to error.

Q: Is there a shortcut for the cross product?

While there are no 'shortcuts' that bypass the math, you can simplify the vectors before taking the cross product if they share a common factor. However, in this case, the numbers are manageable enough to compute directly. Accuracy is your best shortcut!

Question

Let a line pass through two distinct points $P(-2,-1,3)$ and Q, and be parallel to the vector $3\hat{i}+2\hat{j}+2\hat{k}$ If the distance of the point Q from the point $R(1,3,3)$ is 5, then the square of the area of $	riangle PQR$ is equal to:

Options:

(A) 148

(B) 136

(C) 144

(D) 140

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional coordinate space. You have a fixed point $P(-2, -1, 3)$ and a line passing through it, stretching out like a laser beam in the direction of the vector $\vec{d} = 3\hat{i} + 2\hat{j} + 2\hat{k}$.

Somewhere along this infinite line, there is a mysterious point $Q$. We know it sits at a distance of exactly $5$ units from another fixed point $R(1, 3, 3)$.

Our mission is to find the square of the area of the triangle formed by these three points: $P$, $Q$, and $R$.

### The Parametric GPS

To find a point on a line, we use a parameter, $\lambda$. If you start at $P$ and move in the direction of $\vec{d}$, your position at any moment is given by $\vec{r} = \vec{P} + \lambda \vec{d}$.

By substituting the coordinates of $P$ and the components of $\vec{d}$, we get the general coordinates for $Q$:
$$Q = (-2 + 3\lambda, -1 + 2\lambda, 3 + 2\lambda)$$

Every point on that line is now captured by the single variable $\lambda$.

### The Distance Constraint

We apply the constraint that the distance between $Q$ and $R$ is $5$. To simplify, we work with the square of the distance, $QR^2 = 25$.

Plugging our parametric $Q$ and fixed $R$ into the distance formula, we get:
$$(3\lambda - 2 - 1)^2 + (2\lambda - 1 - 3)^2 + (2\lambda + 3 - 3)^2 = 25$$

Simplifying the terms inside the brackets, we obtain:
$$(3\lambda - 3)^2 + (2\lambda - 4)^2 + (2\lambda)^2 = 25$$

Expanding these squares yields:
$$9\lambda^2 - 18\lambda + 9 + 4\lambda^2 - 16\lambda + 16 + 4\lambda^2 = 25$$

Combining like terms, we arrive at:
$$17\lambda^2 - 34\lambda + 25 = 25$$

The constants cancel out, leaving us with $17\lambda^2 - 34\lambda = 0$. Factoring this gives $17\lambda(\lambda - 2) = 0$, which implies $\lambda = 0$ or $\lambda = 2$. Since $P$ and $Q$ must be distinct, we reject $\lambda = 0$ and accept $\lambda = 2$.

### The Final Area Calculation

With $\lambda = 2$, we find the coordinates of $Q$:
$$Q = (-2 + 3(2), -1 + 2(2), 3 + 2(2)) = (4, 3, 7)$$

We define two vectors originating from $P$:
$$\vec{PQ} = Q - P = (4 - (-2), 3 - (-1), 7 - 3) = (6, 4, 4)$$
$$\vec{PR} = R - P = (1 - (-2), 3 - (-1), 3 - 3) = (3, 4, 0)$$

The area of the triangle is given by $\frac{1}{2} | \vec{PQ} 	imes \vec{PR} |$. We calculate the cross product:
$$\vec{PQ} 	imes \vec{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 6 & 4 & 4 \ 3 & 4 & 0 \end{vmatrix} = \hat{i}(0 - 16) - \hat{j}(0 - 12) + \hat{k}(24 - 12) = -16\hat{i} + 12\hat{j} + 12\hat{k}$$

The magnitude squared of this cross product is:
$$| \vec{PQ} 	imes \vec{PR} |^2 = (-16)^2 + 12^2 + 12^2 = 256 + 144 + 144 = 544$$

Finally, the square of the area is:
$$	ext{Area}^2 = \left( \frac{1}{2} ight)^2 	imes | \vec{PQ} 	imes \vec{PR} |^2 = \frac{1}{4} 	imes 544 = 136$$

The square of the area of the triangle is **136**.

Select Answer:

Visualized Solution

Visualizing the Geometry

Parametric Form of Point $Q$

Using the Distance Constraint

Simplifying the Expression

Expanding the Quadratic Terms

Solving for $λ$

Determining the Correct $λ$

Finding Coordinates of $Q$

Defining Vectors $P Q$ and $P R$

Calculating the Cross Product

Finding the Square of the Area

Final Conclusion

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Parametric GPS

The Distance Constraint

The Final Area Calculation

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