Question:
If \(P\) divides the line segment joining the points \(A(1,2,-1)\) and \(B(-1,0,1)\) externally in the ratio \(1:2\) and \(Q=(1,3,-1)\), then \(PQ=\)
If \(P\) divides the line segment joining the points \(A(1,2,-1)\) and \(B(-1,0,1)\) externally in the ratio \(1:2\) and \(Q=(1,3,-1)\), then \(PQ=\)
Show Hint
For external division of points in 3D, carefully use
\[
\left(
\frac{mx_2-nx_1}{m-n},
\frac{my_2-ny_1}{m-n},
\frac{mz_2-nz_1}{m-n}
\right)
\]
and then apply the 3D distance formula.
Updated On: Jun 22, 2026
- \(\sqrt{10}\)
- \(3\)
- \(1\)
- \(\sqrt{13}\)
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Use the section formula for external division.
Given points, \[ A(1,2,-1) \] and \[ B(-1,0,1) \] Point \(P\) divides \(AB\) externally in the ratio \[ 1:2 \] Using the external section formula, \[ P\left( \frac{m x_2-n x_1}{m-n}, \frac{m y_2-n y_1}{m-n}, \frac{m z_2-n z_1}{m-n} \right) \] where \[ m=1,\qquad n=2 \] Substituting the coordinates, \[ x=\frac{1(-1)-2(1)}{1-2} \] \[ x=\frac{-1-2}{-1}=3 \] Similarly, \[ y=\frac{1(0)-2(2)}{1-2} \] \[ y=\frac{-4}{-1}=4 \] Again, \[ z=\frac{1(1)-2(-1)}{1-2} \] \[ z=\frac{1+2}{-1}=-3 \] Hence, \[ P=(3,4,-3) \]
Step 2: Write the coordinates of \(Q\).
Given, \[ Q=(1,3,-1) \]
Step 3: Apply the distance formula.
Distance between \[ P(3,4,-3) \] and \[ Q(1,3,-1) \] is \[ PQ= \sqrt{ (3-1)^2+(4-3)^2+(-3+1)^2 } \] \[ = \sqrt{ 2^2+1^2+(-2)^2 } \] \[ = \sqrt{4+1+4} \] \[ = \sqrt{9} \] \[ =3 \]
Step 4: Final conclusion.
Therefore, \[ \boxed{3} \]
Given points, \[ A(1,2,-1) \] and \[ B(-1,0,1) \] Point \(P\) divides \(AB\) externally in the ratio \[ 1:2 \] Using the external section formula, \[ P\left( \frac{m x_2-n x_1}{m-n}, \frac{m y_2-n y_1}{m-n}, \frac{m z_2-n z_1}{m-n} \right) \] where \[ m=1,\qquad n=2 \] Substituting the coordinates, \[ x=\frac{1(-1)-2(1)}{1-2} \] \[ x=\frac{-1-2}{-1}=3 \] Similarly, \[ y=\frac{1(0)-2(2)}{1-2} \] \[ y=\frac{-4}{-1}=4 \] Again, \[ z=\frac{1(1)-2(-1)}{1-2} \] \[ z=\frac{1+2}{-1}=-3 \] Hence, \[ P=(3,4,-3) \]
Step 2: Write the coordinates of \(Q\).
Given, \[ Q=(1,3,-1) \]
Step 3: Apply the distance formula.
Distance between \[ P(3,4,-3) \] and \[ Q(1,3,-1) \] is \[ PQ= \sqrt{ (3-1)^2+(4-3)^2+(-3+1)^2 } \] \[ = \sqrt{ 2^2+1^2+(-2)^2 } \] \[ = \sqrt{4+1+4} \] \[ = \sqrt{9} \] \[ =3 \]
Step 4: Final conclusion.
Therefore, \[ \boxed{3} \]
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