Alternative English Year - 2017 Full Marks : 100
Class 12 Ncert solutions Maths || 3d Geometry
Class 12 Ncert solutions Maths || 3d Geometry
Exercise 11.2
Exercise 11.3
Exercise 11.1
1. If a line makes angles \(90^{o},135^{o},45^{o}\)with the x, y and z-axes respectively, find its direction cosines.
Sol: Let l, m, n be the direction cosines of the line.
\(\therefore\) l =\(\cos 90^{o}=0\),
m=\(\cos 135^{o}=\cos (90^{o} + 45^{o})=-\sin 45^{0}=-\frac{1}{\sqrt2}\)
and n =\(\cos 45^{o}=\frac{1}{\sqrt2}\)
So, the direction cosines are 0,\(-\frac{1}{\sqrt2}\),\(\frac{1}{\sqrt2}\)
2. Find the direction cosines of a line which makes equal angles with the coordinate axes.
Sol: Let l,m and n be the direction cosines of the line. Since they make equal angles with the coordinate axes so the direction cosines are equal i.e l=m=n=k(say)
We also know that
\begin{align}
l^2 + m^2 + n^2 &= 1\\
k^2 + k^2 + k^2 &= 1\\
3k^2 &= 1\\
k^2 &= \frac{1}{3}\\
k &= \pm \frac{1}{\sqrt3}\\
\end{align}
\(\therefore\) The required direction cosines are \(\pm \frac{1}{\sqrt3},\pm \frac{1}{\sqrt3},\pm \frac{1}{\sqrt3}\)
3. If a line has the direction ratios -18,12,-4, then what are its direction cosines?
Sol: Let l,m,n be the direction cosines.
Then \(\frac{l}{a}=\frac{m}{b}=\frac{n}{c}\)
\(\frac{l}{-18}=\frac{m}{12}=\frac{n}{-4}\)
\(\frac{l}{-9}=\frac{m}{6}=\frac{n}{-2}=k(say)\)
Therefore l= -9k, m = 6k ,n = -2k
\(k=\pm \frac{1}{\sqrt{a^2 + b^2 + c^2}}=\pm \frac{1}{\sqrt{9^2 + 6^2 + 2^2}}=\pm \frac{1}{\sqrt{121}}=\pm \frac{1}{11}\)
Thus, the direction cosines are given by
\(l=-\frac{9}{11},b=\frac{6}{11},n=-\frac{2}{11}\)
4. Show that the points (2,3,4), (-1,-2,1), (5,8,7) are collinear.
Sol: Let the points (2,3,4), (-1,-2,1), (5,8,7) be denoted by A, B and C respectively
Direction ratio of the line AB are
-1-2,-2-3,1-4, i.e. -3,-5,-3
Direction ratio of the line BC are
5+1, 8+2,7-1 i.e. 6,10,6
It is clear that the direction ratios of AB and BC are proportional. Hence, AB is parallel to BC. But the point B is common to both AB and AC. Therefore, A,B,C are collinear points.
5. Find the direction cosines of the sides of the triangle whose vertices are (3,5,-4), (-1,1,2) and (-5,-5,-2).
Sol: Let the vertices (3,5,-4), (-1,1,2) and (-5,-5,-2) be denoted by P, Q and R respectively.
The direction cosines passing through to two points \(P(x_1,y_1,z_1)\) and \(Q(x_2,y_2,z_2)\) are given by
$$\frac{x_2-x_1}{PQ}, \frac{y_2-y_1}{PQ},\frac{z_2-z_1}{PQ} $$
where PQ=\(\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 +(z_2-z_1)^2}\)
The direction cosines of line passing through P(3,5,-4) and (-1,1,2) are
\(\frac{-1- 3}{\sqrt{(-1-3)^2+(1-5)^2+(2+4)^2}}, \frac{1-5}{\sqrt{(-1-3)^2+(1-5)^2+(2+4)^2}},\frac{2+4}{\sqrt{(-1-3)^2+(1-5)^2+(2+4)^2}} \)
$$\frac{-4}{\sqrt{68}},\frac{-4}{\sqrt{68}},\frac{6}{\sqrt{68}}$$
$$\frac{-4}{2\sqrt{17}},\frac{-4}{2\sqrt{17}},\frac{6}{2\sqrt{17}}$$
$$ \frac{-2}{\sqrt{17}},\frac{-2}{\sqrt{17}},\frac{3}{\sqrt{17}}$$
The direction cosines of the line passing through Q(-1, 1, 2) and R(-5, -5, -2) are
\( \frac{-5-+1}{\sqrt{(-5+1)^2+(-5-1)^2+(-2-2)^2}}, \frac{-5-1}{\sqrt{(-5+1)^2+(-5-1)^2+(-2-2)^2}},\frac{-2-2}{\sqrt{(-5+1)^2+(-5-1)^2+(-2-2)^2}}\)
$$ \frac{-2}{\sqrt{17}},\frac{-3}{\sqrt{17}}, \frac{-2}{\sqrt{17}}$$
The direction cosines of the line passing through R(-5, -5, -2) and P(3, 5, -4)
\( \frac{3+ 5}{\sqrt{(-5-3)^2+ (-5-5)^2+ (-2+4)^2}}, \frac{5+ 5}{\sqrt{(-5-3)^2+ (-5-5)^2+ (-2+4)^2}},\frac{4-2}{\sqrt{(-5-3)^2+ (-5-5)^2+ (-2+4)^2}}\)
\(\frac{}{\sqrt{168}},\frac{10}{\sqrt{168}},\frac{-2}{\sqrt{168}}\)
\(\frac{4}{\sqrt{42}},\frac{5}{\sqrt{42}},\frac{-1}{\sqrt{42}}\)
Class 12 Ncert solutions Maths || 3d Geometry
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