This Question Is From Ex6.1
Q3. The D.R of AB are -2,2,1 if A=(4,1,5) and l(AB)=6 units, fin the coordinate of B

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$$\begin{gathered} \mathrm{Let} \mathrm{co}-\mathrm{ordinates} \mathrm{of} \mathrm{B} \mathrm{be} \left(\mathrm{x},\mathrm{y},\mathrm{z}\right) \\ \mathrm{Co}-\mathrm{ordinates} \mathrm{of} \mathrm{A} \mathrm{are} \mathrm{A}\left(4,1,5\right) \\ \mathrm{Direction} \mathrm{ratios} \mathrm{of} \mathrm{AB}=\left(\mathrm{x}-4,\mathrm{y}-1,\mathrm{z}-5\right) \\ \mathrm{Direction} \mathrm{cosine} \mathrm{of} \mathrm{AB} \mathrm{will} \mathrm{be}: \\ \left(\frac{\left(\mathrm{x}-4\right)}{\sqrt{{\left(\mathrm{x}-4\right)}^2+{\left(\mathrm{y}-1\right)}^2+{\left(\mathrm{z}-5\right)}^2}},\frac{\left(\mathrm{y}-1\right)}{\sqrt{{\left(\mathrm{x}-4\right)}^2+{\left(\mathrm{y}-1\right)}^2+{\left(\mathrm{z}-5\right)}^2}},\frac{\left(\mathrm{z}-1\right)}{\sqrt{{\left(\mathrm{x}-4\right)}^2+{\left(\mathrm{y}-1\right)}^2+{\left(\mathrm{z}-5\right)}^2}}\right) \\ \mathrm{As} \mathrm{AB}=5 \mathrm{units} , \mathrm{by} \mathrm{distance} \mathrm{formula}, \mathrm{we} \mathrm{have}, \\ \sqrt{{\left(\mathrm{x}-4\right)}^2+{\left(\mathrm{y}-1\right)}^2+{\left(\mathrm{z}-5\right)}^2}=6 \\ \Rightarrow \mathrm{Direction} \mathrm{cosine} \mathrm{of} \mathrm{AB} \mathrm{will} \mathrm{be}: \\ \left(\frac{\mathrm{x}-4}{6},\frac{\mathrm{y}-1}{6},\frac{\mathrm{z}-1}{6}\right) \\ \mathrm{Given} \mathrm{that} \mathrm{direction} \mathrm{ratios} \mathrm{of} \mathrm{AB}=\left(-2,2,1\right) \\ \Rightarrow \mathrm{Direction} \mathrm{cosine} \mathrm{of} \mathrm{AB} \mathrm{will} \mathrm{be} \left(\frac{-2}{\sqrt{4+4+1}},\frac{2}{\sqrt{4+4+1}},\frac{1}{\sqrt{4+4+1}}\right)=\left(-\frac{2}{3},\frac{2}{3},\frac{1}{3}\right) \\ \Rightarrow \left(\frac{\mathrm{x}-4}{6},\frac{\mathrm{y}-1}{6},\frac{\mathrm{z}-1}{6}\right)=\left(-\frac{2}{3},\frac{2}{3},\frac{1}{3}\right) \\ \Rightarrow \frac{\mathrm{x}-4}{6}=-\frac{2}{3} \\ \Rightarrow 3\mathrm{x}-12=-12 \\ \Rightarrow 3\mathrm{x}=24 \\ \Rightarrow \mathrm{x}=8 \\ \frac{\mathrm{y}-1}{6}=\frac{2}{3} \\ \Rightarrow 3\mathrm{y}-3=12 \\ \Rightarrow 3\mathrm{y}=15 \\ \Rightarrow \mathrm{y}=5 \\ \frac{\mathrm{z}-1}{6}=\frac{1}{3} \\ \Rightarrow 3\mathrm{z}-3=6 \\ \Rightarrow 3\mathrm{z}=9 \\ \Rightarrow \mathrm{z}=3 \\ \mathbf{\Rightarrow }\left(\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\right)\mathbf{=}\left(\mathbf{8}\mathbf{,}\mathbf{5}\mathbf{,}\mathbf{3}\right)\mathbf{ }\left(\mathbf{Answer}\right) \end{gathered}$$

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