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Question: If ABCD is a parallelogram, \(\overset{\rightarrow}{AB} = 2\mathbf{i} + 4\mathbf{j} - 5\mathbf{k}\) ...

If ABCD is a parallelogram, AB→=2i+4j−5k\overset{\rightarrow}{AB} = 2\mathbf{i} + 4\mathbf{j} - 5\mathbf{k} and AD→=i+2j+3k,\overset{\rightarrow}{AD} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k}, then the unit vector in the direction of BD is

A

169(i+2j−8k)\frac{1}{\sqrt{69}}(\mathbf{i} + 2\mathbf{j} - 8\mathbf{k})

B

169(i+2j−8k)\frac{1}{69}(\mathbf{i} + 2\mathbf{j} - 8\mathbf{k})

C

169(−i−2j+8k)\frac{1}{\sqrt{69}}( - \mathbf{i} - 2\mathbf{j} + 8\mathbf{k})

D

169(−i−2j+8k)\frac{1}{69}( - \mathbf{i} - 2\mathbf{j} + 8\mathbf{k})

Answer

169(−i−2j+8k)\frac{1}{\sqrt{69}}( - \mathbf{i} - 2\mathbf{j} + 8\mathbf{k})

Explanation

Solution

Since AB→+BD→=AD→⇒BD→=AD→−AB→\overset{\rightarrow}{AB} + \overset{\rightarrow}{BD} = \overset{\rightarrow}{AD} \Rightarrow \overset{\rightarrow}{BD} = \overset{\rightarrow}{AD} - \overset{\rightarrow}{AB}

=(i+2j+3k)−(2i+4j−5k)=−i−2j+8k= (\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}) - (2\mathbf{i} + 4\mathbf{j} - 5\mathbf{k}) = - \mathbf{i} - 2\mathbf{j} + 8\mathbf{k}

Hence unit vector in the direction of BD→\overset{\rightarrow}{BD} is

−i−2j+8k∣−i−2j+8k∣=−i−2j+8k69.\frac{- \mathbf{i} - 2\mathbf{j} + 8\mathbf{k}}{| - \mathbf{i} - 2\mathbf{j} + 8\mathbf{k}|} = \frac{- \mathbf{i} - 2\mathbf{j} + 8\mathbf{k}}{\sqrt{69}}.