MHT CET · Maths · Vector Algebra
The perimeter of the triangle whose vertices have the position vectos \(\hat{\imath}+\hat{\jmath}+\hat{\mathrm{k}}, 5 \hat{\imath}+3 \hat{\jmath}-3 \hat{\mathrm{k}}\) and \(2 \hat{\imath}+5 \hat{\jmath}+9 \hat{\mathrm{k}}\) is
- A \((\sqrt{15}-\sqrt{157})\) units
- B \((15+\sqrt{157})\) units
- C \((15-\sqrt{157})\) units
- D \((\sqrt{15}+\sqrt{157})\) units
Answer & Solution
Correct Answer
(B) \((15+\sqrt{157})\) units
Step-by-step Solution
Detailed explanation
Let \(\quad \overline{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overline{\mathrm{b}}=5 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}, \overline{\mathrm{c}}=2 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+9 \hat{\mathrm{k}}\)
\(\therefore \overline{\mathrm{AB}}=4 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-4 \hat{\mathrm{k}} \quad \Rightarrow \mid \overline{\mathrm{AB}}=\sqrt{16+4+16}=6\)
\(\overline{\mathrm{BC}}=-3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+12 \hat{\mathrm{k}} \Rightarrow|\overline{\mathrm{BC}}|=\sqrt{9+4+144}=\sqrt{157}\)
\(\overline{\mathrm{AC}}=\hat{\mathrm{i}}+4 \hat{\mathrm{j}}+8 \hat{\mathrm{k}} \quad \Rightarrow \mid \overline{\mathrm{AC}}=\sqrt{1+16+64}=9\)
Perimeter \(=|\overline{\mathrm{AB}}|+|\overline{\mathrm{BC}}|+|\overline{\mathrm{AC}}|=15+\sqrt{157}\)
\(\therefore \overline{\mathrm{AB}}=4 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-4 \hat{\mathrm{k}} \quad \Rightarrow \mid \overline{\mathrm{AB}}=\sqrt{16+4+16}=6\)
\(\overline{\mathrm{BC}}=-3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+12 \hat{\mathrm{k}} \Rightarrow|\overline{\mathrm{BC}}|=\sqrt{9+4+144}=\sqrt{157}\)
\(\overline{\mathrm{AC}}=\hat{\mathrm{i}}+4 \hat{\mathrm{j}}+8 \hat{\mathrm{k}} \quad \Rightarrow \mid \overline{\mathrm{AC}}=\sqrt{1+16+64}=9\)
Perimeter \(=|\overline{\mathrm{AB}}|+|\overline{\mathrm{BC}}|+|\overline{\mathrm{AC}}|=15+\sqrt{157}\)
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