JEE Mains · Maths · STD 12 - 11. three dimension geometry
If the equation of the line passing through the point \(\left(0,-\frac{1}{2}, 0\right)\) and perpendicular to the lines \(\vec{r}=\lambda(\hat{i}+a \hat{j}+b \hat{k})\) and \(\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}-\hat{\mathrm{j}}-6 \hat{\mathrm{k}})+\mu(-b \hat{\mathrm{i}}+\mathrm{a} \hat{\mathrm{j}}+5 \hat{\mathrm{k}})\) is \(\frac{\mathrm{x}-1}{-2}=\frac{\mathrm{y}+4}{\mathrm{~d}}=\frac{\mathrm{z}-\mathrm{c}}{-4}\), then \(\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}\) is equal to:
- A 10
- B 14
- C 13
- D 12
Answer & Solution
Correct Answer
(B) 14
Step-by-step Solution
Detailed explanation
Line is \(\perp^{\mathrm{r}}\) to 2 line \(\Rightarrow\) line will be parallel to \((i+a \hat{j}+b \hat{k}) \times(-b \hat{i}+a \hat{j}+5 \hat{k})\) Parallel vector along the required line is…
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