Skip to content
Exercise 6.1 · Q10

Q.A line passes through (3,−1,2)(3, -1, 2) and is perpendicular to lines rˉ=(i^+j^−k^)+λ(2i^−2j^+k^)\bar{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(2\hat{i} - 2\hat{j} + \hat{k}) and rˉ=(2i^+j^−3k^)+μ(i^−2j^+2k^)\bar{r} = (2\hat{i} + \hat{j} - 3\hat{k}) + \mu(\hat{i} - 2\hat{j} + 2\hat{k}). Find its equation.

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
7% · 10/145 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

The two given lines have direction vectors bˉ=2i^−2j^+k^\bar{b} = 2\hat{i}-2\hat{j}+\hat{k} and cˉ=i^−2j^+2k^\bar{c} = \hat{i}-2\hat{j}+2\hat{k}. Since the required line must be perpendicular to both of these directions, it must run parallel to bˉ×cˉ\bar{b}\times\bar{c}, exactly as in Ex.(3) and Ex.(8) of this section.

Computing the cross product:

bˉ×cˉ=∣i^j^k^2−211−22∣=i^((−2)(2)−(1)(−2))−j^((2)(2)−(1)(1))+k^((2)(−2)−(−2)(1))\bar{b}\times\bar{c} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&-2&1\\1&-2&2\end{vmatrix} = \hat{i}\big((-2)(2)-(1)(-2)\big) - \hat{j}\big((2)(2)-(1)(1)\big) + \hat{k}\big((2)(-2)-(-2)(1)\big)

=i^(−4+2)−j^(4−1)+k^(−4+2)=−2i^−3j^−2k^.= \hat{i}(-4+2) - \hat{j}(4-1) + \hat{k}(-4+2) = -2\hat{i} - 3\hat{j} - 2\hat{k}. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.