Where Mindful Learning Begins
Where Mindful Learning Begins
This Class 12 Maths NCERT Solutions Chapter 11 Miscellaneous Exercise page covers all 5 questions, solved step-by-step. It's a short exercise, but each question is a slightly different spin on the direction-ratio and shortest-distance ideas from Exercises 11.1 and 11.2, closing out the chapter with the kind of question CBSE actually sets rather than a straight repeat of the earlier drills.
Question 1 is a neat algebraic proof — the angle between lines with direction ratios a,b,c and b-c,c-a,a-b turns out to be a right angle no matter what a,b,c actually are, which is exactly the kind of "prove it's always true" question CBSE likes to slip in. Question 2 asks for the equation of a coordinate axis itself, a case students often overthink. Question 3 is a direct application of the perpendicularity condition to solve for an unknown. Questions 4 and 5 return to the shortest-distance and line-construction techniques from Exercise 11.2 — the last question in particular, finding a line perpendicular to two other lines at once, is really a cross-product question wearing a 3D-geometry costume, and previews exactly the kind of reasoning you'll need once planes are introduced.
The angle θ between the lines satisfies \cos\theta = \left|\dfrac{a(b-c)+b(c-a)+c(a-b)}{\sqrt{a^2+b^2+c^2}\sqrt{(b-c)^2+(c-a)^2+(a-b)^2}}\right|.
Expanding the numerator: a(b-c)+b(c-a)+c(a-b) = ab-ac+bc-ab+ac-bc = 0.
Since the numerator is identically 0 for every choice of a, b, c, \cos\theta=0 always.
A line parallel to the x-axis has direction ratios (1, 0, 0), and it passes through the origin (0, 0, 0).
Cartesian equation: \dfrac{x-0}{1} = \dfrac{y-0}{0} = \dfrac{z-0}{0}, i.e. \dfrac{x}{1}=\dfrac{y}{0}=\dfrac{z}{0} — this represents the line y=0,\ z=0, which is the x-axis itself.
Vector equation: \vec{r} = \lambda\hat{i}.
Direction ratios: line 1 is (-3, 2k, 2); line 2 is (3k, 1, -5).
Perpendicularity requires (-3)(3k)+(2k)(1)+(2)(-5) = 0.
-9k+2k-10=0 \Rightarrow -7k=10.
\vec{a_1}=6\hat{i}+2\hat{j}+2\hat{k}, \vec{b_1}=\hat{i}-2\hat{j}+2\hat{k}; \vec{a_2}=-4\hat{i}-\hat{k}, \vec{b_2}=3\hat{i}-2\hat{j}-2\hat{k}.
\vec{a_2}-\vec{a_1} = -10\hat{i}-2\hat{j}-3\hat{k}.
\vec{b_1}\times\vec{b_2} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&-2&2\\3&-2&-2\end{vmatrix} = \hat{i}(4+4)-\hat{j}(-2-6)+\hat{k}(-2+6) = 8\hat{i}+8\hat{j}+4\hat{k}.
|\vec{b_1}\times\vec{b_2}| = \sqrt{64+64+16} = \sqrt{144} = 12.
(\vec{b_1}\times\vec{b_2})\cdot(\vec{a_2}-\vec{a_1}) = 8(-10)+8(-2)+4(-3) = -80-16-12 = -108.
d = \dfrac{|-108|}{12} = 9.
The two given lines have direction ratios \vec{u}=3\hat{i}-16\hat{j}+7\hat{k} and \vec{v}=3\hat{i}+8\hat{j}-5\hat{k}.
A line perpendicular to both must be along \vec{u}\times\vec{v}.
\vec{u}\times\vec{v} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\3&-16&7\\3&8&-5\end{vmatrix} = \hat{i}(80-56)-\hat{j}(-15-21)+\hat{k}(24+48)
= 24\hat{i}+36\hat{j}+72\hat{k} = 12(2\hat{i}+3\hat{j}+6\hat{k}).
So the required line has direction ratios (2,3,6) and passes through (1,2,-4).
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