Exercise 11.1 Practice
Chapter 11: Three Dimensional Geometry – NCERT Solutions
Q1
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If a line makes angles $90^\circ, 135^\circ, 45^\circ$ with the $x, y$ and $z$-axes respectively, find its direction cosines.
Let direction cosines be $l, m, n$.
$l = \cos 90^\circ = 0$
$m = \cos 135^\circ = \cos(180^\circ - 45^\circ) = -\cos 45^\circ = -\frac{1}{\sqrt{2}}$
$n = \cos 45^\circ = \frac{1}{\sqrt{2}}$
$\boxed{0, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}}$
Q2
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Find the direction cosines of a line which makes equal angles with the coordinate axes.
Let the angles be $\alpha, \beta, \gamma$. Given $\alpha = \beta = \gamma$.
Direction cosines are $l = \cos \alpha, m = \cos \alpha, n = \cos \alpha$.
We know $l^2 + m^2 + n^2 = 1 \Rightarrow 3\cos^2 \alpha = 1 \Rightarrow \cos^2 \alpha = \frac{1}{3}$.
$\cos \alpha = \pm \frac{1}{\sqrt{3}}$.
$\boxed{\pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}}$
Q3
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If a line has the direction ratios $-18, 12, -4$, then what are its direction cosines?
Direction ratios are $a = -18, b = 12, c = -4$.
Magnitude $r = \sqrt{(-18)^2 + (12)^2 + (-4)^2} = \sqrt{324 + 144 + 16} = \sqrt{484} = 22$.
Direction cosines are $\frac{a}{r}, \frac{b}{r}, \frac{c}{r}$.
$l = \frac{-18}{22} = \frac{-9}{11}, \quad m = \frac{12}{22} = \frac{6}{11}, \quad n = \frac{-4}{22} = \frac{-2}{11}$.
$\boxed{-\frac{9}{11}, \frac{6}{11}, -\frac{2}{11}}$
Q4
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Show that the points $(2, 3, 4), (-1, -2, 1), (5, 8, 7)$ are collinear.
Let points be $A(2, 3, 4), B(-1, -2, 1), C(5, 8, 7)$.
Direction ratios of AB: $(-1-2, -2-3, 1-4) = (-3, -5, -3)$.
Direction ratios of BC: $(5-(-1), 8-(-2), 7-1) = (6, 10, 6)$.
Since $\frac{6}{-3} = \frac{10}{-5} = \frac{6}{-3} = -2$, the direction ratios are proportional.
Thus AB is parallel to BC. Since B is a common point, A, B, C are collinear.
$\boxed{\text{Shown}}$
Q5
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Find the direction cosines of the sides of the triangle whose vertices are $(3, 5, -4), (-1, 1, 2)$ and $(-5, -5, -2)$.
Vertices: $A(3, 5, -4), B(-1, 1, 2), C(-5, -5, -2)$.
Side AB: DRs $(-4, -4, 6)$. Mag $\sqrt{16+16+36} = \sqrt{68} = 2\sqrt{17}$. DCs: $\frac{-2}{\sqrt{17}}, \frac{-2}{\sqrt{17}}, \frac{3}{\sqrt{17}}$.
Side BC: DRs $(-4, -6, -4)$. Mag $\sqrt{16+36+16} = \sqrt{68} = 2\sqrt{17}$. DCs: $\frac{-2}{\sqrt{17}}, \frac{-3}{\sqrt{17}}, \frac{-2}{\sqrt{17}}$.
Side CA: DRs $(8, 10, -2)$. Mag $\sqrt{64+100+4} = \sqrt{168} = 2\sqrt{42}$. DCs: $\frac{4}{\sqrt{42}}, \frac{5}{\sqrt{42}}, \frac{-1}{\sqrt{42}}$.
$\boxed{\text{AB: } \frac{-2}{\sqrt{17}}, \frac{-2}{\sqrt{17}}, \frac{3}{\sqrt{17}}; \text{ BC: } \frac{-2}{\sqrt{17}}, \frac{-3}{\sqrt{17}}, \frac{-2}{\sqrt{17}}; \text{ CA: } \frac{4}{\sqrt{42}}, \frac{5}{\sqrt{42}}, \frac{-1}{\sqrt{42}}}$