Question

The distance of the point (1, 1, 9) from the point of intersection of the line $\frac{x-3}{1} = \frac{y-4}{2} = \frac{z-5}{2}$ and the plane $x + y + z = 17$ is :

• A. $2\sqrt{19}$
• B. $19\sqrt{2}$
• C. $\sqrt{38}$
• D. 38

Hint:

Always use the parametric form of a line when looking for its intersection with a plane.

Answer 1

The Correct Option is C

To find the distance of the point \( (1, 1, 9) \) from the point of intersection of the given line and plane, we must first determine the point of intersection.

The line is given in the symmetric form:

\(\frac{x-3}{1} = \frac{y-4}{2} = \frac{z-5}{2}\)

Let t be the parameter, then:

• x = 3 + t
• y = 4 + 2t
• z = 5 + 2t

The plane equation is:

x + y + z = 17

Substitute the parametric equations into the plane's equation:

(3 + t) + (4 + 2t) + (5 + 2t) = 17

Simplify to find t:

3 + t + 4 + 2t + 5 + 2t = 17

12 + 5t = 17

5t = 5

t = 1

Substitute t = 1 back into the parametric equations to find the intersection point:

• x = 3 + 1 = 4
• y = 4 + 2 \times 1 = 6
• z = 5 + 2 \times 1 = 7

Thus, the point of intersection is \( (4, 6, 7) \).

Next, calculate the distance between the points \( (1, 1, 9) \) and \( (4, 6, 7) \) using the distance formula:

d = \sqrt{(4 - 1)^2 + (6 - 1)^2 + (7 - 9)^2}

d = \sqrt{3^2 + 5^2 + (-2)^2}

d = \sqrt{9 + 25 + 4}

d = \sqrt{38}

Thus, the distance is \( \sqrt{38} \). Therefore, the correct answer is \( \sqrt{38} \).