The points A(4,5,10),B(2,3,4) and C(1,2,−1) are three vertices of a parallelogram ABCD. Find the vector equations of the sides AB and BC and also find the coordinates of point D.

Engineering

Mathematics

Equation of Straight Line in 3D

Section Formulae and Centres of a Triangle

Locus of a Point

Question

The points A(4,5,10),B(2,3,4) and C(1,2,−1) are three vertices of a parallelogram ABCD. Find the vector equations of the sides AB and BC and also find the coordinates of point D.

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

The points A(4,5,10), B(2,3,4) and C(1,2,–1) are three vertices of parallelogram ABCD.

Let coordinates of D be (x, y, z)
Direction vector along AB is

\vec{a} = (2 - 4) \hat{i} + (3 - 5) \hat{j} + (4 - 10) \hat{k} = - 2 \hat{i} - 2 \hat{j} - 6 \hat{k}

∴ Equation of line AB, is given by

\vec{b} = (4 \hat{i} + 5 \hat{j} + 10 \hat{k}) + λ (2 \hat{i} + 2 \hat{j} + 6 \hat{k})

Direction vector along BC is

\vec{c} = (1 - 2) \hat{i} + (2 - 3) \hat{j} + (- 1 - 4) \hat{k} = - \hat{i} - \hat{j} - 5 \hat{k}

∴ Equation of a line BC, is given by

\vec{d} = (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) + μ (\hat{i} + \hat{j} + 5 \hat{k})

Since ABCD is a parallelogram AC and BD bisect each other
$∴ [\frac{4 + 1}{2}, \frac{5 + 2}{2}, \frac{10 - 1}{2}] = [\frac{2 + x}{2}, \frac{3 + y}{2}, \frac{4 + z}{2}]$

⇒ 2 + x = 5, 3 + y = 7, 4 + z = 9
⇒ x = 3, y = 4, z = 5

Coordinates of D are (3,4,5).