Vector Algebra
Position Vectors
Grade 12
Question:
<p>If ABCD is a parallelogram and the position vectors of A, B and C are $\vec{i} + 3\vec{j} + 5\vec{k}$, $\vec{i} + \vec{j} + \vec{k}$ and $7\vec{i} + 7\vec{j} + 7\vec{k}$, then the position vector of D will be</p>
<p>(a) $7\vec{i} + 5\vec{j} + 3\vec{k}$</p>
<p>(b) $7\vec{i} + 9\vec{j} + 11\vec{k}$</p>
<p>(c) $9\vec{i} + 11\vec{j} + 13\vec{k}$</p>
<p>(d) $8\vec{i} + 8\vec{j} + 8\vec{k}$</p>
Step-by-Step Solution
Key Concept: In a parallelogram, opposite sides are equal vectors. Use $\vec{AB} = \vec{DC}$ to find the fourth vertex.
Solution: Let position vector of D is $x\hat{i} + y\hat{j} + z\hat{k}$, then $\vec{AB} = \vec{DC}$. $\vec{AB} = (\hat{i} + \hat{j} + \hat{k}) - (\hat{i} + 3\hat{j} + 5\hat{k}) = -2\hat{j} - 4\hat{k}$ $\vec{DC} = (7\hat{i} + 7\hat{j} + 7\hat{k}) - (x\hat{i} + y\hat{j} + z\hat{k}) = (7-x)\hat{i} + (7-y)\hat{j} + (7-z)\hat{k}$ Equating: $-2\hat{j} - 4\hat{k} = (7-x)\hat{i} + (7-y)\hat{j} + (7-z)\hat{k}$ Therefore $x = 7, y = 9$ and $z = 11$ Position vector of D = $7\hat{i} + 9\hat{j} + 11\hat{k}$
Correct Answer: b