Question

Let \((\alpha,\beta,\gamma)\) be the foot of the perpendicular from the point \((25,2,41)\) on the line \[ \frac{x-4}{3}=\frac{y+1}{7}=\frac{z-2}{3}. \] Then \(\alpha+\beta+\gamma\) is equal to:

• A. \(41\)
• B. \(42\)
• C. \(44\)
• D. \(45\)

Hint:

For foot of perpendicular problems in 3D, always use the dot product condition with the direction vector of the line.

Answer 1

The Correct Option is C

Concept:
A point on a line in symmetric form can be written using a parameter.
The foot of the perpendicular from a point to a line satisfies the condition that the vector joining them is perpendicular to the direction vector of the line.
Two vectors are perpendicular if their dot product is zero.
Step 1: Parametric form of the given line Let \[ \frac{x-4}{3}=\frac{y+1}{7}=\frac{z-2}{3}=t \] Then a general point on the line is: \[ A(4+3t,\,-1+7t,\,2+3t) \] The direction vector of the line is: \[ \vec{d}=(3,7,3) \]
Step 2: Use perpendicularity condition Given point \(P(25,2,41)\). For \(A\) to be the foot of the perpendicular, \[ \overrightarrow{AP}\cdot\vec{d}=0 \] \[ (25-(4+3t),\;2-(-1+7t),\;41-(2+3t))\cdot(3,7,3)=0 \] \[ (21-3t,\;3-7t,\;39-3t)\cdot(3,7,3)=0 \]
Step 3: Solve for \(t\) \[ 3(21-3t)+7(3-7t)+3(39-3t)=0 \] \[ 63-9t+21-49t+117-9t=0 \] \[ 201-67t=0 \Rightarrow t=3 \]
Step 4: Find the foot of the perpendicular Substitute \(t=3\): \[ \alpha=4+9=13,\quad \beta=-1+21=20,\quad \gamma=2+9=11 \]
Step 5: Required sum \[ \alpha+\beta+\gamma=13+20+11=44 \]