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From the points P(3, -1,11), a perpendic...

From the points `P(3, -1,11)`, a perpendicular is drawn on the line L given by the equation `x/2=(y-2)/3=(z-3)/4`. Let Q be the foot of the perpendicular.
What are the drection ratios of the line segment PQ?

A

`(: 1,6,4 :)`

B

`(: -1,6,-4 :)`

C

`(: -1,-6,4 :)`

D

`(: 2,-6,4 :)`

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The correct Answer is:
To find the direction ratios of the line segment PQ, where P is the point (3, -1, 11) and Q is the foot of the perpendicular from P to the line L given by the equations \( \frac{x}{2} = \frac{y-2}{3} = \frac{z-3}{4} \), we can follow these steps: ### Step 1: Identify the direction ratios of the line L The line L can be expressed in parametric form. From the equation \( \frac{x}{2} = \frac{y-2}{3} = \frac{z-3}{4} \), we can set a parameter \( t \): - \( x = 2t \) - \( y = 3t + 2 \) - \( z = 4t + 3 \) The direction ratios of the line L are (2, 3, 4). ### Step 2: Find the coordinates of point Q Let Q be the point on the line L corresponding to the parameter \( t \). Thus, the coordinates of Q are: - \( Q(2t, 3t + 2, 4t + 3) \) ### Step 3: Find the vector PQ The vector PQ can be expressed as: \[ PQ = Q - P = (2t - 3, (3t + 2) - (-1), (4t + 3) - 11) \] This simplifies to: \[ PQ = (2t - 3, 3t + 3, 4t - 8) \] ### Step 4: Find the condition for PQ to be perpendicular to L For PQ to be perpendicular to the line L, the dot product of PQ and the direction ratios of the line L must equal zero: \[ PQ \cdot (2, 3, 4) = 0 \] Calculating the dot product: \[ (2t - 3) \cdot 2 + (3t + 3) \cdot 3 + (4t - 8) \cdot 4 = 0 \] Expanding this gives: \[ 4t - 6 + 9t + 9 + 16t - 32 = 0 \] Combining like terms: \[ 29t - 29 = 0 \] Solving for \( t \): \[ t = 1 \] ### Step 5: Substitute \( t \) back to find Q Now substitute \( t = 1 \) back into the parametric equations for Q: - \( Q(2 \cdot 1, 3 \cdot 1 + 2, 4 \cdot 1 + 3) = Q(2, 5, 7) \) ### Step 6: Calculate the direction ratios of PQ Now we can find the direction ratios of the line segment PQ: \[ PQ = Q - P = (2 - 3, 5 - (-1), 7 - 11) = (-1, 6, -4) \] ### Conclusion The direction ratios of the line segment PQ are \((-1, 6, -4)\).

To find the direction ratios of the line segment PQ, where P is the point (3, -1, 11) and Q is the foot of the perpendicular from P to the line L given by the equations \( \frac{x}{2} = \frac{y-2}{3} = \frac{z-3}{4} \), we can follow these steps: ### Step 1: Identify the direction ratios of the line L The line L can be expressed in parametric form. From the equation \( \frac{x}{2} = \frac{y-2}{3} = \frac{z-3}{4} \), we can set a parameter \( t \): - \( x = 2t \) - \( y = 3t + 2 \) - \( z = 4t + 3 \) ...
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