TS EAMCET · Maths · Three Dimensional Geometry
If the shortest distance between the lines \(\mathbf{r}=(3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})+t(-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}})\) and \(\mathbf{r}=(\hat{\mathbf{i}}-7 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})+s(\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})\) is equivalent to projection of \(\mathbf{P}=-2 \hat{\mathbf{i}}+11 \hat{\mathbf{j}}\) on \(\mathbf{Q}\) then a possible vector \(\mathbf{Q}\) is
- A \(\hat{i}+5 \hat{j}-3 \hat{k}\)
- B \(5 \hat{i}-\hat{j}+3 \hat{k}\)
- C \(\hat{\mathrm{i}}+3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}\)
- D \(3 \hat{i}+5 \hat{j}-\hat{k}\)
Answer & Solution
Correct Answer
(C) \(\hat{\mathrm{i}}+3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}\)
Step-by-step Solution
Detailed explanation
The shortest distance between given lines \(\mathbf{r}=(3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})+t(-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}})\) and…
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