Let $P$ be the image of the point $Q(7, -2, 5)$ in the line $L: \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}$ and $R(5, p, q)$ be a point on $L$. Then the square of the area of $\triangle PQR$ is:

Step 1: Find the coordinates of point R

The given line $L$ is in symmetric form. We can express the coordinates of any point on the line in terms of a parameter $\lambda$. Since $R(5, p, q)$ lies on the line, we can substitute its $x$ -coordinate to find $\lambda$ and then determine $p$ and $q$.

Step 2: Find the foot of the perpendicular from Q to L

Let $M$ be the foot of the perpendicular from $Q$ to the line $L$. Any point on $L$ can be written as $(2\lambda + 1, 3\lambda - 1, 4\lambda)$. The vector $\vec{QM}$ is perpendicular to the direction vector of the line $L$, which is $\vec{b} = (2, 3, 4)$. Their dot product must be zero.

Step 3: Determine the coordinates of M and P

Solving the dot product equation for $\lambda$, we find $\lambda = 1$. Substituting this value back into the general point on the line gives the coordinates of $M$. Since $P$ is the image of $Q$ in the line $L$, $M$ is the midpoint of $PQ$. We can use the midpoint formula to find the coordinates of $P$.

Step 4: Calculate the area of triangle PQR

Now that we have the coordinates of $P$, $Q$, and $R$, we can calculate the area of $\triangle PQR$. The area of a triangle with vertices $P, Q, R$ can be found using half the magnitude of the cross product of two vectors forming two sides of the triangle, for example, $\vec{RP}$ and $\vec{RQ}$.

Step 5: Compute the cross product and its magnitude

We compute the cross product of $\vec{RP}$ and $\vec{RQ}$. Then, we find the magnitude of this resultant vector. This magnitude is twice the area of the triangle.

Step 6: Calculate the square of the area

Finally, we calculate the area of the triangle and then square the result to get the required answer.

More from this chapter

• Let ABCD be a tetrahedron such that the edges AB, AC and AD are mutually perpendicular. Let the areas of the triangles ABC, ACD and ADB be 5, 6 and 7 square uni
• Let L_1: (x-1)/(1) = (y-2)/(1) = (z-1)/(2) and L_2: (x+1)/(1) = (y-2)/(2) = (z)/(4) be two lines. Let L_3 be a line passing through the point (α, β,) and be per
• The line L_1 is parallel to the vector a = -3 i + 2 j + 4 k and passes through the point (7, 6, 2) and the line L_2 is parallel to the vector b = 2 i + j + 3 k
• The distance of the line (x-2)/(2)=(y-6)/(3)=(z-3)/(4) from the point (1,4,0) along the line (x)/(4)=(y-2)/(2)=(z+3)/(3) is:
• Find the shortest distance between the skew lines: r⃗₁ = (i + j + k) + λ(2i + 3j + 4k) and r⃗₂ = (2i + 3j + 5k) + μ(i + 2j + 3k).
• Find the equation of the plane passing through the point (1, 2, -3) and perpendicular to the planes x + 2y + 3z = 4 and 2x - 3y + 4z = 5.
• Let the line passing through the points (-1, 2, 1) and parallel to the line (x-1)/(2) = (y+1)/(3) = (z)/(4) intersect the line (x+2)/(3) = (y-3)/(2) = (z-4)/(3)
• Let the shortest distance between the lines (x-3)/(3) = (y-α)/(-1) = (z-3)/(1) and (x+3)/(-3) = (y+7)/(2) = (z-β)/(4) be 3√30. Then the positive value of 5α + β

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