NCERT Solutions: Class 12 Maths Chapter 11 - Three Dimensional Geometry

The direction cosines are $\cos90^\circ,\cos135^\circ,\cos45^\circ$. Hence they are $0,-\dfrac1{\sqrt2},\dfrac1{\sqrt2}$.

Let the direction cosines be $l,m,n$. Since the line makes equal angles with the coordinate axes, $l=m=n$. Also $l^2+m^2+n^2=1$. Thus $3l^2=1$, so $l=m=n=\dfrac1{\sqrt3}$ for one direction; reversing the direction changes all signs.

For direction ratios $a,b,c$, the corresponding direction cosines are $\dfrac{a}{\sqrt{a^2+b^2+c^2}},\dfrac{b}{\sqrt{a^2+b^2+c^2}},\dfrac{c}{\sqrt{a^2+b^2+c^2}}$. Here $\sqrt{(-18)^2+12^2+(-4)^2}=\sqrt{484}=22$. Hence the direction cosines are $-\dfrac{18}{22},\dfrac{12}{22},-\dfrac4{22}$, i.e. $-\dfrac9{11},\dfrac6{11},-\dfrac2{11}$.

Let $A(2,3,4)$, $B(-1,-2,1)$ and $C(5,8,7)$. Direction ratios of $AB$ are $(-1-2,-2-3,1-4)=(-3,-5,-3)$. Direction ratios of $AC$ are $(5-2,8-3,7-4)=(3,5,3)=-1(-3,-5,-3)$. Since the direction ratios are proportional, $AB$ and $AC$ are the same line through $A$. Hence the points are collinear.

Let $A(3,5,-4)$, $B(-1,1,2)$ and $C(-5,-5,-2)$. For $AB$, direction ratios are $(-4,-4,6)$ and length is $\sqrt{16+16+36}=2\sqrt{17}$, so the direction cosines are $-\dfrac2{\sqrt{17}},-\dfrac2{\sqrt{17}},\dfrac3{\sqrt{17}}$. For $BC$, direction ratios are $(-4,-6,-4)$ and length is $2\sqrt{17}$, giving $-\dfrac2{\sqrt{17}},-\dfrac3{\sqrt{17}},-\dfrac2{\sqrt{17}}$. For $CA$, direction ratios are $(8,10,-2)$ and length is $2\sqrt{42}$, giving $\dfrac4{\sqrt{42}},\dfrac5{\sqrt{42}},-\dfrac1{\sqrt{42}}$.

Let the direction cosine triples be $\left(\dfrac{12}{13},-\dfrac3{13},-\dfrac4{13}\right)$, $\left(\dfrac4{13},\dfrac{12}{13},\dfrac3{13}\right)$ and $\left(\dfrac3{13},-\dfrac4{13},\dfrac{12}{13}\right)$. Their pairwise dot products are $\dfrac{48-36-12}{169}=0$, $\dfrac{36+12-48}{169}=0$ and $\dfrac{12-48+36}{169}=0$. Hence each pair of lines is perpendicular.

Direction ratios of the first line are $(3-1,4+1,-2-2)=(2,5,-4)$. Direction ratios of the second line are $(3-0,5-3,6-2)=(3,2,4)$. Their dot product is $2\cdot3+5\cdot2+(-4)\cdot4=6+10-16=0$. Hence the lines are perpendicular.

Direction ratios of the first line are $(2-4,3-7,4-8)=(-2,-4,-4)$. Direction ratios of the second line are $(1+1,2+2,5-1)=(2,4,4)=-1(-2,-4,-4)$. Since the direction ratios are proportional, the lines are parallel.

A line through point vector $\hat{i}+2\hat{j}+3\hat{k}$ and parallel to $3\hat{i}+2\hat{j}-2\hat{k}$ is $\vec r=\hat{i}+2\hat{j}+3\hat{k}+\lambda(3\hat{i}+2\hat{j}-2\hat{k})$. Equating components gives $x=1+3\lambda$, $y=2+2\lambda$, $z=3-2\lambda$, hence $\dfrac{x-1}{3}=\dfrac{y-2}{2}=\dfrac{z-3}{-2}$.

The required line passes through $(2,-1,4)$ and has direction ratios $(1,2,-1)$. Thus $\vec r=2\hat{i}-\hat{j}+4\hat{k}+\lambda(\hat{i}+2\hat{j}-\hat{k})$. In coordinates, $x=2+\lambda$, $y=-1+2\lambda$, $z=4-\lambda$, giving $\dfrac{x-2}{1}=\dfrac{y+1}{2}=\dfrac{z-4}{-1}$.

The given line has direction ratios $(3,5,6)$. A parallel line through $(-2,4,-5)$ has the same direction ratios. Hence its cartesian equation is $\dfrac{x-(-2)}{3}=\dfrac{y-4}{5}=\dfrac{z-(-5)}{6}$, i.e. $\dfrac{x+2}{3}=\dfrac{y-4}{5}=\dfrac{z+5}{6}$.

The line passes through $(5,-4,6)$ and has direction ratios $(3,7,2)$. Therefore its vector form is $\vec r=5\hat{i}-4\hat{j}+6\hat{k}+\lambda(3\hat{i}+7\hat{j}+2\hat{k})$.

The angle between two lines depends on their direction vectors. (i) The direction vectors are $(3,2,6)$ and $(1,2,2)$. Thus $\cos\theta=\dfrac{3\cdot1+2\cdot2+6\cdot2}{\sqrt{3^2+2^2+6^2}\sqrt{1^2+2^2+2^2}}=\dfrac{19}{7\cdot3}=\dfrac{19}{21}$. (ii) The direction vectors are $(1,-1,-2)$ and $(3,-5,-4)$. Thus $\cos\theta=\dfrac{1\cdot3+(-1)(-5)+(-2)(-4)}{\sqrt6\sqrt{50}}=\dfrac{16}{10\sqrt3}=\dfrac{8\sqrt3}{15}$.

(i) The direction ratios are $(2,5,-3)$ and $(-1,8,4)$. Hence $\cos\theta=\dfrac{|2(-1)+5(8)+(-3)(4)|}{\sqrt{2^2+5^2+(-3)^2}\sqrt{(-1)^2+8^2+4^2}}=\dfrac{26}{9\sqrt{38}}$. (ii) The direction ratios are $(2,2,1)$ and $(4,1,8)$. Hence $\cos\theta=\dfrac{2\cdot4+2\cdot1+1\cdot8}{\sqrt{4+4+1}\sqrt{16+1+64}}=\dfrac{18}{27}=\dfrac23$.

For the first line, taking the common parameter as $\lambda$ gives direction ratios $\left(-3,\dfrac{2p}{7},2\right)$. For the second line, the direction ratios are $\left(-\dfrac{3p}{7},1,-5\right)$. Perpendicular lines have zero dot product, so $(-3)\left(-\dfrac{3p}{7}\right)+\dfrac{2p}{7}-10=0$. Hence $\dfrac{11p}{7}=10$, giving $p=\dfrac{70}{11}$.

The direction ratios of the two lines are $(7,-5,1)$ and $(1,2,3)$. Their dot product is $7\cdot1+(-5)\cdot2+1\cdot3=7-10+3=0$. Hence the lines are perpendicular.

Here $\vec a_1=(1,2,1)$, $\vec b_1=(1,-1,1)$, $\vec a_2=(2,-1,-1)$ and $\vec b_2=(2,1,2)$. The shortest distance is $\dfrac{|(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)|}{|\vec b_1\times\vec b_2|}$. Now $\vec b_1\times\vec b_2=(-3,0,3)$ and $\vec a_2-\vec a_1=(1,-3,-2)$. Thus the distance is $\dfrac{|(1,-3,-2)\cdot(-3,0,3)|}{\sqrt{9+9}}=\dfrac9{3\sqrt2}=\dfrac3{\sqrt2}$.

For the two lines, take $\vec a_1=(-1,-1,-1)$, $\vec b_1=(7,-6,1)$, $\vec a_2=(3,5,7)$ and $\vec b_2=(1,-2,1)$. Now $\vec b_1\times\vec b_2=(-4,-6,-8)$ and $|\vec b_1\times\vec b_2|=2\sqrt{29}$. Also $\vec a_2-\vec a_1=(4,6,8)$ and $|(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)|=116$. Hence the shortest distance is $\dfrac{116}{2\sqrt{29}}=2\sqrt{29}$.

Here $\vec a_1=(1,2,3)$, $\vec b_1=(1,-3,2)$, $\vec a_2=(4,5,6)$ and $\vec b_2=(2,3,1)$. Now $\vec b_1\times\vec b_2=(-9,3,9)$ and $|\vec b_1\times\vec b_2|=3\sqrt{19}$. Also $\vec a_2-\vec a_1=(3,3,3)$ and $|(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)|=9$. Therefore the shortest distance is $\dfrac9{3\sqrt{19}}=\dfrac3{\sqrt{19}}$.

The first line may be written with $\vec a_1=(1,-2,3)$ and $\vec b_1=(-1,1,-2)$. The second has $\vec a_2=(1,-1,-1)$ and $\vec b_2=(1,2,-2)$. Now $\vec b_1\times\vec b_2=(2,-4,-3)$, so $|\vec b_1\times\vec b_2|=\sqrt{29}$. Also $\vec a_2-\vec a_1=(0,1,-4)$, and $|(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)|=8$. Hence the shortest distance is $\dfrac8{\sqrt{29}}$.

Let the direction ratios be $(a,b,c)$ and $(b-c,c-a,a-b)$. Their dot product is $a(b-c)+b(c-a)+c(a-b)=ab-ac+bc-ab+ac-bc=0$. Since the dot product is zero, the angle between the lines is $90^\circ$.

A line parallel to the $x$-axis has direction ratios $(1,0,0)$. Since it passes through the origin, its vector equation is $\vec r=\lambda\hat{i}$. In cartesian form, points on it have arbitrary $x$ but $y=0$ and $z=0$.

The direction ratios of the two lines are $(-3,2k,2)$ and $(3k,1,-5)$. Since the lines are perpendicular, their dot product is zero: $(-3)(3k)+(2k)(1)+2(-5)=0$. Hence $-9k+2k-10=0$, so $-7k=10$ and $k=-\dfrac{10}{7}$.

Here $\vec a_1=(6,2,2)$, $\vec b_1=(1,-2,2)$, $\vec a_2=(-4,0,-1)$ and $\vec b_2=(3,-2,-2)$. The shortest distance is $\dfrac{|(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)|}{|\vec b_1\times\vec b_2|}$. Now $\vec b_1\times\vec b_2=(8,8,4)$, whose magnitude is $12$. Also $\vec a_2-\vec a_1=(-10,-2,-3)$, and the absolute scalar triple product is $108$. Hence the shortest distance is $\dfrac{108}{12}=9$.

The direction ratios of the given lines are $(3,-16,7)$ and $(3,8,-5)$. A line perpendicular to both must be parallel to their cross product: $(3,-16,7)\times(3,8,-5)=(24,36,72)=12(2,3,6)$. Therefore the required line through $(1,2,-4)$ is $\vec r=\hat{i}+2\hat{j}-4\hat{k}+\lambda(2\hat{i}+3\hat{j}+6\hat{k})$.