Let PQR be a triangle with R(–1, 4, 2). Suppose M(2, 1, 2) is the mid point of PQ. The distance of the centroid of from the point of intersection of the lines and is [2024]

Question

Let PQR be a triangle with R(–1, 4, 2). Suppose M(2, 1, 2) is the mid point of PQ. The distance of the centroid of $△ P Q R$ from the point of intersection of the lines $\frac{x - 2}{0} = \frac{y}{2} = \frac{z + 3}{- 1}$ and $\frac{x - 1}{1} = \frac{y + 3}{- 3} = \frac{z + 1}{1}$ is [2024]

Smart Achievers · Accepted Answer

(1)

Let centroid G divides MR in the ratio 1 : 2.

So, centroid is given by

$∴ G (\frac{4 - 1}{3}, \frac{2 + 4}{3}, \frac{4 + 2}{3})$ i.e., G(1, 2, 2)

Let $l_{1} : \frac{x - 2}{0} = \frac{y}{2} = \frac{z + 3}{- 1} = λ$ (say)

$\Rightarrow x = 2, y = 2 λ, z = - λ - 3$

$l_{2} : \frac{x - 1}{1} = \frac{y + 3}{- 3} = \frac{z + 1}{1} = r$ (say)

$\Rightarrow$ x = r + 1, y = –3r – 3, z = r – 1

Now, $2 = r + 1 \Rightarrow r = 1$ and $2 λ = - 3 r - 3 \Rightarrow λ = - 3$ .

$∴$ Point of intersection of lines $l_{1}$ and $l_{2}$ is given by A(2, –6, 0).

$∴$ Required distance, $A G = \sqrt{1 + 64 + 4} = \sqrt{69}$ .

Solution

Q.
Let PQR be a triangle with R(–1, 4, 2). Suppose M(2, 1, 2) is the mid point of PQ. The distance of the centroid of $△ P Q R$ from the point of intersection of the lines $\frac{x - 2}{0} = \frac{y}{2} = \frac{z + 3}{- 1}$ and $\frac{x - 1}{1} = \frac{y + 3}{- 3} = \frac{z + 1}{1}$ is [2024]

1 $\sqrt{69}$

2 69

3 $\sqrt{99}$

4 9

Want Us to Email you About Special Offers & Updates?

Solution

Q. Let PQR be a triangle with R(–1, 4, 2). Suppose M(2, 1, 2) is the mid point of PQ. The distance of the centroid of △PQR from the point of intersection of the lines x–20=y2=z+3–1 and x–11=y+3–3=z+11 is [2024]

1 69

2 69

3 99

4 9

Want Us to Email you About Special Offers & Updates?

Q.
Let PQR be a triangle with R(–1, 4, 2). Suppose M(2, 1, 2) is the mid point of PQ. The distance of the centroid of $△ P Q R$ from the point of intersection of the lines $\frac{x - 2}{0} = \frac{y}{2} = \frac{z + 3}{- 1}$ and $\frac{x - 1}{1} = \frac{y + 3}{- 3} = \frac{z + 1}{1}$ is [2024]

1 $\sqrt{69}$

3 $\sqrt{99}$