MHT CET · Maths · Three Dimensional Geometry
If \(\triangle \mathrm{ABC}\) is right angled at \(\mathrm{A}\), where \(\mathrm{A} \equiv(4,2, x), \mathrm{B} \equiv(3,1,8)\) and \(\mathrm{C} \equiv(2,-1,2)\), then the value of \(x\) is
- A \(4\)
- B \(2\)
- C \(3\)
- D \(1\)
Answer & Solution
Correct Answer
(D) \(1\)
Step-by-step Solution
Detailed explanation
Since \(\triangle \mathrm{ABC}\) is right angled at \(\mathrm{A}\),
\(\overline{\mathrm{AB}} \cdot \overline{\mathrm{AC}}=0\)
\(\Rightarrow[-\hat{\mathrm{i}}-\hat{\mathrm{j}}+(8-x) \hat{\mathrm{k}}] \cdot(-2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+(2-x) \hat{\mathrm{k}})=0\)
\(\Rightarrow 2+3+(8-x)(2-x)=0\)
\(\Rightarrow x^2-10 x+21=0\)
\(\Rightarrow(x-3)(x-7)=0\)
\(\Rightarrow x=3 \text { or} x=7\)
\(\overline{\mathrm{AB}} \cdot \overline{\mathrm{AC}}=0\)
\(\Rightarrow[-\hat{\mathrm{i}}-\hat{\mathrm{j}}+(8-x) \hat{\mathrm{k}}] \cdot(-2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+(2-x) \hat{\mathrm{k}})=0\)
\(\Rightarrow 2+3+(8-x)(2-x)=0\)
\(\Rightarrow x^2-10 x+21=0\)
\(\Rightarrow(x-3)(x-7)=0\)
\(\Rightarrow x=3 \text { or} x=7\)
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