NCERT Solutions · Class 12 Mathematics Three Dimensional Geometry

In the symmetric form the denominators are the direction ratios, and two lines are perpendicular when \[a_1a_2+b_1b_2+c_1c_2=0 . \]For \(\displaystyle \dfrac{x-5}{7}=\dfrac{y+2}{-5}=\dfrac{z}{1}\): \(\displaystyle a_1,b_1,c_1=7,\,-5,\,1\).For \(\displaystyle \dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{3}\): \(\displaystyle a_2,b_2,c_2=1,\,2,\,3\).Then \[a_1a_2+b_1b_2+c_1c_2=(7)(1)+(-5)(2)+(1)(3)=7-10+3=0 . \]The sum is zero, so the two lines are perpendicular to each other.

For two skew lines \(\displaystyle \vec{r}=\vec{a_1}+\lambda\vec{b_1}\) and \(\displaystyle \vec{r}=\vec{a_2}+\mu\vec{b_2}\), the shortest distance is \[d=\left|\frac{\left(\vec{b_1}\times\vec{b_2}\right)\cdot\left(\vec{a_2}-\vec{a_1}\right)}{\left|\vec{b_1}\times\vec{b_2}\right|}\right| . \]Here \[\vec{a_1}=\hat{i}+2\hat{j}+\hat{k},\quad \vec{b_1}=\hat{i}-\hat{j}+\hat{k},\qquad \vec{a_2}=2\hat{i}-\hat{j}-\hat{k},\quad \vec{b_2}=2\hat{i}+\hat{j}+2\hat{k}. \]First \(\displaystyle \vec{a_2}-\vec{a_1}=(2-1)\hat{i}+(-1-2)\hat{j}+(-1-1)\hat{k}=\hat{i}-3\hat{j}-2\hat{k}\).Next \[\vec{b_1}\times\vec{b_2}=\left|\begin{array}{rrr} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 2 & 1 & 2 \end{array}\right| =\hat{i}(-2-1)-\hat{j}(2-2)+\hat{k}(1+2)=-3\hat{i}+0\hat{j}+3\hat{k}. \] This is not the zero vector, so the lines are not parallel and the formula applies. \[\left|\vec{b_1}\times\vec{b_2}\right|=\sqrt{9+0+9}=\sqrt{18}=3\sqrt{2}. \]Now the numerator: \[\left(\vec{b_1}\times\vec{b_2}\right)\cdot\left(\vec{a_2}-\vec{a_1}\right)=(-3)(1)+(0)(-3)+(3)(-2)=-3+0-6=-9. \]Hence \[d=\frac{\left|-9\right|}{3\sqrt{2}}=\frac{9}{3\sqrt{2}}=\frac{3}{\sqrt{2}}=\frac{3\sqrt{2}}{2}. \]The shortest distance is \(\displaystyle \dfrac{3\sqrt{2}}{2}\) units.

Put each line in the form \(\displaystyle \vec{r}=\vec{a}+\lambda\vec{b}\) and use \[d=\left|\frac{\left(\vec{b_1}\times\vec{b_2}\right)\cdot\left(\vec{a_2}-\vec{a_1}\right)}{\left|\vec{b_1}\times\vec{b_2}\right|}\right| . \]From \(\displaystyle \dfrac{x+1}{7}=\dfrac{y+1}{-6}=\dfrac{z+1}{1}\): the point is \(\displaystyle (-1,-1,-1)\), so \(\displaystyle \vec{a_1}=-\hat{i}-\hat{j}-\hat{k}\), and \(\displaystyle \vec{b_1}=7\hat{i}-6\hat{j}+\hat{k}\).From \(\displaystyle \dfrac{x-3}{1}=\dfrac{y-5}{-2}=\dfrac{z-7}{1}\): \(\displaystyle \vec{a_2}=3\hat{i}+5\hat{j}+7\hat{k}\) and \(\displaystyle \vec{b_2}=\hat{i}-2\hat{j}+\hat{k}\).Subtract, watching the double negatives: \[\vec{a_2}-\vec{a_1}=\left(3-(-1)\right)\hat{i}+\left(5-(-1)\right)\hat{j}+\left(7-(-1)\right)\hat{k}=4\hat{i}+6\hat{j}+8\hat{k}. \]Cross product: \[\vec{b_1}\times\vec{b_2}=\left|\begin{array}{rrr} \hat{i} & \hat{j} & \hat{k} \\ 7 & -6 & 1 \\ 1 & -2 & 1 \end{array}\right| =\hat{i}(-6+2)-\hat{j}(7-1)+\hat{k}(-14+6)=-4\hat{i}-6\hat{j}-8\hat{k}, \] \[\left|\vec{b_1}\times\vec{b_2}\right|=\sqrt{16+36+64}=\sqrt{116}=2\sqrt{29}. \]Numerator: \[\left(\vec{b_1}\times\vec{b_2}\right)\cdot\left(\vec{a_2}-\vec{a_1}\right)=(-4)(4)+(-6)(6)+(-8)(8)=-16-36-64=-116. \]Hence \[d=\frac{116}{2\sqrt{29}}=\frac{58}{\sqrt{29}}=\frac{58\sqrt{29}}{29}=2\sqrt{29}. \]The shortest distance is \(\displaystyle 2\sqrt{29}\) units.

Both lines are already in the form \(\displaystyle \vec{r}=\vec{a}+\lambda\vec{b}\), so use \[d=\left|\frac{\left(\vec{b_1}\times\vec{b_2}\right)\cdot\left(\vec{a_2}-\vec{a_1}\right)}{\left|\vec{b_1}\times\vec{b_2}\right|}\right| . \]Read off \[\vec{a_1}=\hat{i}+2\hat{j}+3\hat{k},\quad \vec{b_1}=\hat{i}-3\hat{j}+2\hat{k},\qquad \vec{a_2}=4\hat{i}+5\hat{j}+6\hat{k},\quad \vec{b_2}=2\hat{i}+3\hat{j}+\hat{k}, \] \[\vec{a_2}-\vec{a_1}=3\hat{i}+3\hat{j}+3\hat{k}. \]Cross product: \[\vec{b_1}\times\vec{b_2}=\left|\begin{array}{rrr} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ 2 & 3 & 1 \end{array}\right| =\hat{i}(-3-6)-\hat{j}(1-4)+\hat{k}(3+6)=-9\hat{i}+3\hat{j}+9\hat{k}, \] \[\left|\vec{b_1}\times\vec{b_2}\right|=\sqrt{81+9+81}=\sqrt{171}=3\sqrt{19}. \]Numerator: \[\left(\vec{b_1}\times\vec{b_2}\right)\cdot\left(\vec{a_2}-\vec{a_1}\right)=(-9)(3)+(3)(3)+(9)(3)=-27+9+27=9. \]Hence \[d=\frac{9}{3\sqrt{19}}=\frac{3}{\sqrt{19}}=\frac{3\sqrt{19}}{19}. \]The shortest distance is \(\displaystyle \dfrac{3}{\sqrt{19}}\) units.

The equations are given in expanded parametric form, so first collect the constant part and the part carrying the parameter to get each into the form \(\displaystyle \vec{r}=\vec{a}+t\,\vec{b}\).First line: \[\vec{r}=(1-t)\hat{i}+(t-2)\hat{j}+(3-2t)\hat{k}=\left(\hat{i}-2\hat{j}+3\hat{k}\right)+t\left(-\hat{i}+\hat{j}-2\hat{k}\right), \] so \(\displaystyle \vec{a_1}=\hat{i}-2\hat{j}+3\hat{k}\) and \(\displaystyle \vec{b_1}=-\hat{i}+\hat{j}-2\hat{k}\).Second line (the whole \(\displaystyle \hat{k}\) bracket is negated, so its constant term is \(\displaystyle -1\) and its coefficient of \(\displaystyle s\) is \(\displaystyle -2\)): \[\vec{r}=(s+1)\hat{i}+(2s-1)\hat{j}-(2s+1)\hat{k}=\left(\hat{i}-\hat{j}-\hat{k}\right)+s\left(\hat{i}+2\hat{j}-2\hat{k}\right), \] so \(\displaystyle \vec{a_2}=\hat{i}-\hat{j}-\hat{k}\) and \(\displaystyle \vec{b_2}=\hat{i}+2\hat{j}-2\hat{k}\).Then \[\vec{a_2}-\vec{a_1}=(1-1)\hat{i}+(-1+2)\hat{j}+(-1-3)\hat{k}=0\hat{i}+\hat{j}-4\hat{k}. \]Cross product: \[\vec{b_1}\times\vec{b_2}=\left|\begin{array}{rrr} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & -2 \\ 1 & 2 & -2 \end{array}\right| =\hat{i}(-2+4)-\hat{j}(2+2)+\hat{k}(-2-1)=2\hat{i}-4\hat{j}-3\hat{k}, \] \[\left|\vec{b_1}\times\vec{b_2}\right|=\sqrt{4+16+9}=\sqrt{29}. \]Apply \(\displaystyle d=\left|\dfrac{\left(\vec{b_1}\times\vec{b_2}\right)\cdot\left(\vec{a_2}-\vec{a_1}\right)}{\left|\vec{b_1}\times\vec{b_2}\right|}\right|\): \[\left(\vec{b_1}\times\vec{b_2}\right)\cdot\left(\vec{a_2}-\vec{a_1}\right)=(2)(0)+(-4)(1)+(-3)(-4)=0-4+12=8, \] \[d=\frac{8}{\sqrt{29}}=\frac{8\sqrt{29}}{29}. \]The shortest distance is \(\displaystyle \dfrac{8}{\sqrt{29}}\) units.