NCERT Solutions · Class 12 Mathematics Three Dimensional Geometry

If two lines have direction ratios \(\displaystyle a_1, b_1, c_1\) and \(\displaystyle a_2, b_2, c_2\), the angle \(\displaystyle \theta\) between them satisfies \[\cos\theta = \frac{a_1a_2 + b_1b_2 + c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\;\sqrt{a_2^2+b_2^2+c_2^2}} . \] Here \(\displaystyle (a_1, b_1, c_1) = (a,\, b,\, c)\) and \(\displaystyle (a_2, b_2, c_2) = (b-c,\, c-a,\, a-b)\).Evaluate the numerator alone: \[a(b-c) + b(c-a) + c(a-b) = ab - ac + bc - ab + ca - cb = 0 , \] since the six terms cancel in pairs (\(\displaystyle ab\) with \(\displaystyle -ab\), \(\displaystyle -ac\) with \(\displaystyle ca\), \(\displaystyle bc\) with \(\displaystyle -cb\)). This is an identity in \(\displaystyle a, b, c\) — no special values are needed.Both denominators are non-zero (direction ratios of a line are never all zero), so \[\cos\theta = \frac{0}{\sqrt{a^2+b^2+c^2}\;\sqrt{(b-c)^2+(c-a)^2+(a-b)^2}} = 0 . \]Hence \[\theta = \frac{\pi}{2} = 90^\circ , \] that is, the two lines are perpendicular for every choice of \(\displaystyle a, b, c\).

A line through the point with position vector \(\displaystyle \vec{a}\) and parallel to the vector \(\displaystyle \vec{b}\) has vector equation \(\displaystyle \vec{r} = \vec{a} + \lambda\vec{b}\), \(\displaystyle \lambda \in \mathbb{R}\).The line passes through the origin, so \(\displaystyle \vec{a} = \vec{0}\). It is parallel to the \(\displaystyle x\)-axis, whose direction is \(\displaystyle \hat{i}\); so take \(\displaystyle \vec{b} = \hat{i}\), i.e. direction ratios \(\displaystyle 1, 0, 0\). Therefore \[\vec{r} = \vec{0} + \lambda\,\hat{i} = \lambda\,\hat{i} . \]In Cartesian form, using \(\displaystyle (x_1, y_1, z_1) = (0,0,0)\) and direction ratios \(\displaystyle 1, 0, 0\), \[\frac{x-0}{1} = \frac{y-0}{0} = \frac{z-0}{0} . \] The zero denominators are the standard symbolic convention here: they are read as saying that those coordinates cannot change, i.e. \[y = 0, \qquad z = 0 , \] with \(\displaystyle x\) free. The line is the \(\displaystyle x\)-axis itself.

Two lines \(\displaystyle \dfrac{x-x_1}{a_1} = \dfrac{y-y_1}{b_1} = \dfrac{z-z_1}{c_1}\) and \(\displaystyle \dfrac{x-x_2}{a_2} = \dfrac{y-y_2}{b_2} = \dfrac{z-z_2}{c_2}\) are perpendicular if and only if their direction ratios satisfy \[a_1a_2 + b_1b_2 + c_1c_2 = 0 . \] (The points on the lines play no part — perpendicularity of lines in space is a condition on directions only.)Read the direction ratios off the denominators: \[(a_1, b_1, c_1) = (-3,\; 2k,\; 2), \qquad (a_2, b_2, c_2) = (3k,\; 1,\; -5). \]Impose the condition: \[(-3)(3k) + (2k)(1) + (2)(-5) = 0 \] \[-9k + 2k - 10 = 0 \] \[-7k = 10 . \]Hence \[k = -\frac{10}{7} . \]

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 \[d = \left| \frac{(\vec{b}_1 \times \vec{b}_2)\cdot(\vec{a}_2 - \vec{a}_1)}{|\vec{b}_1 \times \vec{b}_2|} \right| . \]From the given equations, \[\vec{a}_1 = 6\hat{i} + 2\hat{j} + 2\hat{k}, \qquad \vec{b}_1 = \hat{i} - 2\hat{j} + 2\hat{k}, \] \[\vec{a}_2 = -4\hat{i} + 0\,\hat{j} - \hat{k}, \qquad \vec{b}_2 = 3\hat{i} - 2\hat{j} - 2\hat{k} . \] Note the missing \(\displaystyle \hat{j}\) in the second line's point: its \(\displaystyle y\)-coordinate is \(\displaystyle 0\), not \(\displaystyle 1\).The common perpendicular direction: \[\vec{b}_1 \times \vec{b}_2 = \left|\begin{array}{rrr} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 2 \\ 3 & -2 & -2 \end{array}\right| = \hat{i}(4+4) - \hat{j}(-2-6) + \hat{k}(-2+6) = 8\hat{i} + 8\hat{j} + 4\hat{k} . \] Since \(\displaystyle \vec{b}_1 \times \vec{b}_2 \neq \vec{0}\), the lines are not parallel, so this formula applies. Its magnitude: \[|\vec{b}_1 \times \vec{b}_2| = \sqrt{8^2 + 8^2 + 4^2} = \sqrt{64+64+16} = \sqrt{144} = 12 . \]The joining vector: \[\vec{a}_2 - \vec{a}_1 = (-4-6)\hat{i} + (0-2)\hat{j} + (-1-2)\hat{k} = -10\hat{i} - 2\hat{j} - 3\hat{k} . \]Scalar triple product: \[(\vec{b}_1 \times \vec{b}_2)\cdot(\vec{a}_2 - \vec{a}_1) = 8(-10) + 8(-2) + 4(-3) = -80 - 16 - 12 = -108 . \] This is non-zero, confirming the lines are skew (not intersecting).Therefore \[d = \frac{|-108|}{12} = 9 . \]The shortest distance between the two lines is \(\displaystyle 9\) units.

A line perpendicular to two given lines must be perpendicular to both of their direction vectors, so its direction is along the cross product of those two vectors.The given lines have direction vectors read off the denominators: \[\vec{b}_1 = 3\hat{i} - 16\hat{j} + 7\hat{k}, \qquad \vec{b}_2 = 3\hat{i} + 8\hat{j} - 5\hat{k} . \]Required direction: \[\vec{b} = \vec{b}_1 \times \vec{b}_2 = \left|\begin{array}{rrr} \hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{array}\right| \] \[= \hat{i}\big[(-16)(-5) - (7)(8)\big] - \hat{j}\big[(3)(-5) - (7)(3)\big] + \hat{k}\big[(3)(8) - (-16)(3)\big] \] \[= \hat{i}(80 - 56) - \hat{j}(-15 - 21) + \hat{k}(24 + 48) = 24\hat{i} + 36\hat{j} + 72\hat{k} . \] Watch the sign on the middle term: \(\displaystyle -\hat{j}(-36) = +36\hat{j}\).Only the direction matters, so divide by the common factor \(\displaystyle 12\): \[\vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k} . \]The line passes through \(\displaystyle (1, 2, -4)\), i.e. \(\displaystyle \vec{a} = \hat{i} + 2\hat{j} - 4\hat{k}\). Using \(\displaystyle \vec{r} = \vec{a} + \lambda\vec{b}\), \[\vec{r} = \big(\hat{i} + 2\hat{j} - 4\hat{k}\big) + \lambda\big(2\hat{i} + 3\hat{j} + 6\hat{k}\big), \qquad \lambda \in \mathbb{R} . \]Equivalently, in Cartesian form, \[\frac{x-1}{2} = \frac{y-2}{3} = \frac{z+4}{6} . \]