AP EAMCET · Maths · Three Dimensional Geometry
The point of intersection of the lines represented by \(\overline{\mathrm{r}}=(\overline{\mathrm{i}}-6 \overline{\mathrm{j}}+2 \overline{\mathrm{k}})+\mathrm{t}(\overline{\mathrm{i}}+2 \overline{\mathrm{j}}+\overline{\mathrm{k}})\) and \(\overline{\mathrm{r}}=(4 \overline{\mathrm{j}}+\overline{\mathrm{k}})+\mathrm{s}(2 \overline{\mathrm{i}}+\overline{\mathrm{j}}+2 \overline{\mathrm{k}})\) is
- A \(8 \overline{\mathrm{i}}+9 \overline{\mathrm{j}}+10 \overline{\mathrm{k}}\)
- B \(8 \overline{\mathrm{i}}+8 \overline{\mathrm{j}}+7 \overline{\mathrm{k}}\)
- C \(8 \overline{\mathrm{i}}+9 \overline{\mathrm{j}}+8 \overline{\mathrm{k}}\)
- D \(8 \overline{\mathrm{i}}+8 \overline{\mathrm{j}}+9 \overline{\mathrm{k}}\)
Answer & Solution
Correct Answer
(D) \(8 \overline{\mathrm{i}}+8 \overline{\mathrm{j}}+9 \overline{\mathrm{k}}\)
Step-by-step Solution
Detailed explanation
Equating components: \(1 + t = 2s\) \(-6 + 2t = 4 + s\) \(2 + t = 1 + 2s\) From first equation, \(t = 2s - 1\). Substitute \(t\) into second equation: \(-6 + 2(2s - 1) = 4 + s\) \(4s - 8 = 4 + s\) \(3s = 12 \implies s = 4\) Substitute \(s=4\) into \(t = 2s - 1\):…
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