CBSE Class 12 Mathematics HOTs Three Dimensional Geometry Set 02

Very Short Answer Type Questions

Question. Find the coordinates of the point where the line \(\frac{x+3}{3} = \frac{y-1}{-1} = \frac{z-5}{-5}\) cuts the \(XY\)-plane.
Answer: The coordinates of any point on the line \(\frac{x+3}{3} = \frac{y-1}{-1} = \frac{z-5}{-5}\) are given by \((3\lambda - 3, -\lambda + 1, -5\lambda + 5)\).
If \((3\lambda - 3, -\lambda + 1, -5\lambda + 5)\) lies on \(XY\)-plane i.e. \(z = 0\), then:
\(-5\lambda + 5 = 0 \Rightarrow \lambda = 1\)
Hence, the coordinates of the points are:
\((3 \times 1 - 3, -1 \times 1 + 1, -5 \times 1 + 5) = (0, 0, 0)\).

Question. If a line makes angle \(60^\circ\), \(135^\circ\) and \(45^\circ\) with the \(X, Y\) and \(Z\)-axes, respectively. Find its direction cosines.
Answer: Let direction cosines of the line be \(l, m\) and \(n\).
Given, \(\alpha = 60^\circ\), \(\beta = 135^\circ\) and \(\gamma = 45^\circ\)
Then, \(l = \cos\alpha = \cos 60^\circ = \frac{1}{2}\)
\(m = \cos\beta = \cos 135^\circ = -\frac{1}{\sqrt{2}}\)
and \(n = \cos\gamma = \cos 45^\circ = \frac{1}{\sqrt{2}}\)
Hence, the direction cosines of the line are \(\frac{1}{2}\), \(-\frac{1}{\sqrt{2}}\) and \(\frac{1}{\sqrt{2}}\).

Question. If a line makes angles \(60^\circ\), \(45^\circ\) and \(\theta\) with \(X, Y\) and \(Z\)-axes respectively, then find the angle \(\theta\), where \(\theta\) is acute angle.
Answer: Given, \(\alpha = 60^\circ\), \(\beta = 45^\circ\) and \(\gamma = \theta\)
We know that \(l^2 + m^2 + n^2 = 1\)
\(\Rightarrow \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1\)
\(\Rightarrow \cos^2 60^\circ + \cos^2 45^\circ + \cos^2\theta = 1\)
\(\Rightarrow \frac{1}{4} + \frac{1}{2} + \cos^2\theta = 1 \Rightarrow \cos^2\theta = 1 - \frac{3}{4}\)
\(\Rightarrow \cos^2\theta = \frac{1}{4} \Rightarrow \cos\theta = \pm \frac{1}{2}\)
\(\Rightarrow \cos\theta = \frac{1}{2} = \cos 60^\circ\)
or \(\cos\theta = \frac{-1}{2} = \cos 120^\circ\)
\(\Rightarrow \theta = 60^\circ\) or \(120^\circ\)
\(\therefore \theta = 60^\circ\) [\(\because \theta\) is an acute angle]

Question. If a line has direction ratios \(1, -2, 1\), then determine its direction cosines.
Answer: Given direction ratios are \((1, -2, 1)\), i.e. \(a = 1, b = -2\) and \(c = 1\).
Then, \(\sqrt{a^2 + b^2 + c^2} = \sqrt{(1)^2 + (-2)^2 + (1)^2} = \sqrt{6}\)
Now, direction cosines are:
\(l = \frac{a}{\sqrt{a^2+b^2+c^2}} = \frac{1}{\sqrt{6}}\)
\(m = \frac{b}{\sqrt{a^2+b^2+c^2}} = \frac{-2}{\sqrt{6}}\)
and \(n = \frac{c}{\sqrt{a^2+b^2+c^2}} = \frac{1}{\sqrt{6}}\)
Hence, the direction cosines are \(\frac{1}{\sqrt{6}}\), \(\frac{-2}{\sqrt{6}}\), \(\frac{1}{\sqrt{6}}\).

Question. Find the direction ratios of the line segment joining the points \(A(1, 5, 4)\) and \(B(4, 1, -2)\).
Answer: Let \(A(x_1, y_1, z_1) = (1, 5, 4)\) and \(B(x_2, y_2, z_2) = (4, 1, -2)\).
Then, direction ratios of line \(AB\) are \((4 - 1, 1 - 5, -2 - 4)\) i.e. \((3, -4, -6)\).

Question. Check the following points are collinear or not: \(A(5, 7, 8)\), \(B(-1, -2, 1)\) and \(C(2, 3, 4)\).
Answer: Given points are \(A(5, 7, 8)\), \(B(-1, -2, 1)\) and \(C(2, 3, 4)\).
Direction ratios of line joining \(A\) and \(B\) are \((-1 - 5, -2 - 7, 1 - 8)\) i.e. \((-6, -9, -7)\).
Direction ratios of line joining \(B\) and \(C\) are \((2 - (-1), 3 - (-2), 4 - 1)\) i.e. \((3, 5, 3)\).
Now, ratios of direction ratios of \(AB\) and \(BC\) are \(\frac{-6}{3}, \frac{-9}{5}, \frac{-7}{3}\).
Thus, the direction ratios of \(AB\) and \(BC\) are not proportional.
Hence, \(A, B\) and \(C\) are not collinear points.

Question. If the cartesian equation of a line is \(\frac{3-x}{5} = \frac{y+4}{7} = \frac{2z-6}{4}\), then write the vector equation of the line.
Answer: Given, cartesian equation of a line is \(\frac{3-x}{5} = \frac{y+4}{7} = \frac{2z-6}{4}\).
On rewriting the given equation in standard form, we get:
\(\frac{x-3}{-5} = \frac{y+4}{7} = \frac{z-3}{2} = \lambda\) (let)
\(\Rightarrow x = -5\lambda + 3\), \(y = 7\lambda - 4\) and \(z = 2\lambda + 3\)
Now, \(x\hat{i} + y\hat{j} + z\hat{k} = (-5\lambda + 3)\hat{i} + (7\lambda - 4)\hat{j} + (2\lambda + 3)\hat{k}\)
\(\therefore \vec{r} = (3\hat{i} - 4\hat{j} + 3\hat{k}) + \lambda(-5\hat{i} + 7\hat{j} + 2\hat{k})\), which is the required equation of line in vector form.

Question. Write the equation of the straight line through the point \((\alpha, \beta, \gamma)\) and parallel to \(Z\)-axis.
Answer: The vector equation of a line parallel to \(Z\)-axis is \(\vec{m} = 0\hat{i} + 0\hat{j} + \hat{k}\).
Then, the required line passes through the point \(A(\alpha, \beta, \gamma)\), whose position vector is \(\vec{r}_1 = \alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}\) and is parallel to the vector \(\vec{m} = (0\hat{i} + 0\hat{j} + \hat{k})\).
\(\therefore\) The equation is \(\vec{r} = \vec{r}_1 + \lambda\vec{m}\)
\(= (\alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}) + \lambda(0\hat{i} + 0\hat{j} + \hat{k})\)
\(= (\alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}) + \lambda(\hat{k})\).

Question. Find the direction cosines of the line \(\frac{4-x}{2} = \frac{y}{6} = \frac{1-z}{3}\).
Answer: Given equation of line is \(\frac{4-x}{2} = \frac{y}{6} = \frac{1-z}{3}\).
It can be rewritten in standard form as:
\(\frac{x-4}{-2} = \frac{y}{6} = \frac{z-1}{-3}\)
Here, direction ratios of the line are \((-2, 6, -3)\).
\(\therefore\) Direction cosines of the line are:
\(\frac{-2}{\sqrt{(-2)^2 + (6)^2 + (-3)^2}}\), \(\frac{6}{\sqrt{(-2)^2 + (6)^2 + (-3)^2}}\) and \(\frac{-3}{\sqrt{(-2)^2 + (6)^2 + (-3)^2}}\) i.e., \(\frac{-2}{\sqrt{49}}\), \(\frac{6}{\sqrt{49}}\) and \(\frac{-3}{\sqrt{49}}\).
Thus, direction cosines of the line are \(\left(-\frac{2}{7}, \frac{6}{7}, -\frac{3}{7}\right)\).

Question. Write the vector equation of a line passing through the point \((1, -1, 2)\) and parallel to the line whose equation is \(\frac{x-3}{1} = \frac{y-1}{2} = \frac{z+1}{-2}\).
Answer: We know that the vector equation of a line passing through a point with position vector \(\vec{a}\) and parallel to a given vector \(\vec{b}\) is \(\vec{r} = \vec{a} + \lambda\vec{b}\), where \(\lambda \in \mathbb{R}\).
Here, \(\vec{a} = \hat{i} - \hat{j} + 2\hat{k}\) and \(\vec{b} = \hat{i} + 2\hat{j} - 2\hat{k}\).
[\(\because\) direction ratios of given line are \(1, 2\) and \(-2\)]
\(\therefore\) Required vector equation of line is \(\vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \lambda(\hat{i} + 2\hat{j} - 2\hat{k})\), where \(\lambda \in \mathbb{R}\).

Question. Find the cartesian equation of the line which passes through the point \((-2, 4, -5)\) and is parallel to the line \(\frac{x+3}{3} = \frac{4-y}{5} = \frac{z+8}{6}\).
Answer: Given, the required line is parallel to the line:
\(\frac{x+3}{3} = \frac{4-y}{5} = \frac{z+8}{6}\) or \(\frac{x+3}{3} = \frac{y-4}{-5} = \frac{z+8}{6}\).
\(\therefore\) Direction ratios of both lines are proportional to each other.
The required equation of the line passing through \((-2, 4, -5)\) having direction ratios \((3, -5, 6)\) is \(\frac{x+2}{3} = \frac{y-4}{-5} = \frac{z+5}{6}\).

Question. Find the equation of the line which passes through the point \((2, 4, 6)\) and is parallel to the vector \(3\hat{i} + 2\hat{j} - 2\hat{k}\).
Answer: Let \(\vec{a}\) be the position vector of the point \((2, 4, 6)\).
Then, \(\vec{a} = 2\hat{i} + 4\hat{j} + 6\hat{k}\).
Now, the equation of the line passing through the point having position vector \(\vec{a}\) and parallel to \(\vec{b} = 3\hat{i} + 2\hat{j} - 2\hat{k}\) is:
\(\vec{r} = \vec{a} + \lambda\vec{b}\)
\(\Rightarrow \vec{r} = (2\hat{i} + 4\hat{j} + 6\hat{k}) + \lambda(3\hat{i} + 2\hat{j} - 2\hat{k})\)

Question. Find the equation of a line passing through the point \((-3, 2, -4)\) and equally inclined to the axes.
Answer: We have, point \(A(-3, 2, -4)\).
According to the question, \(l = m = n\).
Therefore, required equation of line is \(\frac{x+3}{1} = \frac{y-2}{1} = \frac{z+4}{1} \Rightarrow x+3 = y-2 = z+4\).

Question. Find the angle between two lines whose direction ratios are \((2, 1, 2)\) and \((4, 8, 1)\).
Answer: Let \(\theta\) be the angle between the given lines.
Here, \(\vec{b}_1 = 2\hat{i} + \hat{j} + 2\hat{k}\) and \(\vec{b}_2 = 4\hat{i} + 8\hat{j} + \hat{k}\).
\(\therefore \cos\theta = \left|\frac{\vec{b}_1 \cdot \vec{b}_2}{|\vec{b}_1||\vec{b}_2|}\right| = \left|\frac{(2\hat{i} + \hat{j} + 2\hat{k}) \cdot (4\hat{i} + 8\hat{j} + \hat{k})}{|2\hat{i} + \hat{j} + 2\hat{k}||4\hat{i} + 8\hat{j} + \hat{k}|}\right|\)
\(= \frac{8+8+2}{\sqrt{4+1+4}\cdot\sqrt{16+64+1}} = \frac{18}{3 \times 9} = \frac{2}{3}\)
\(\Rightarrow \theta = \cos^{-1}\left(\frac{2}{3}\right)\).

Question. Show that the equations of line \(\frac{x-2}{2} = \frac{y-1}{7} = \frac{z+3}{-3}\) and \(\frac{x+2}{-1} = \frac{y-8}{2} = \frac{z-5}{4}\) are perpendicular to each other.
Answer: Given lines are \(\frac{x-2}{2} = \frac{y-1}{7} = \frac{z+3}{-3}\) and \(\frac{x+2}{-1} = \frac{y-8}{2} = \frac{z-5}{4}\).
We know that if two lines are perpendicular, then:
\(a_1a_2 + b_1b_2 + c_1c_2 = 0\)
\(\therefore a_1a_2 + b_1b_2 + c_1c_2 = 2 \times (-1) + 7 \times 2 + (-3) \times 4\)
\(= -2 + 14 - 12 = 0\)
Hence, the given lines are perpendicular to each other.

Short Answer Type Questions

Question. A line makes the same angle \(\theta\) with each of the \(X\) and \(Z\)-axes. If the angle \(\beta\), which it makes with \(Y\)-axis, is such that \(\sin^2\beta = 3\sin^2\theta\), then find the value of \(\cos^2\theta\).
Answer: Clearly, \(\cos^2\theta + \cos^2\beta + \cos^2\theta = 1\)
\(\Rightarrow 2\cos^2\theta + 1 - \sin^2\beta = 1\)
\(\Rightarrow 2\cos^2\theta - \sin^2\beta = 0\)
\(\Rightarrow 2\cos^2\theta - 3\sin^2\theta = 0\) [\(\because \sin^2\beta = 3\sin^2\theta\) (given)]
\(\Rightarrow \tan^2\theta = \frac{2}{3}\)
\(\therefore \cos^2\theta = \frac{1}{1 + \tan^2\theta} = \frac{1}{1 + \frac{2}{3}} = \frac{3}{5}\).

Question. The equations of a line is \(5x-3 = 15y+7 = 3-10z\). Write the direction cosines of the line.
Answer: Given equation of a line is:
\(5x-3 = 15y+7 = 3-10z\) ...(i)
Let us first convert the equation in standard form:
\(\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}\) ...(ii)
Let us divide Eq. (i) by LCM (coefficients of \(x, y\) and \(z\)) i.e. LCM \((5, 15, 10) = 30\).
Now, the Eq. (i) becomes:
\(\frac{5x-3}{30} = \frac{15y+7}{30} = \frac{3-10z}{30}\)
\(\Rightarrow \frac{5\left(x - \frac{3}{5}\right)}{30} = \frac{15\left(y + \frac{7}{15}\right)}{30} = \frac{-10\left(z - \frac{3}{10}\right)}{30}\)
\(\Rightarrow \frac{x - \frac{3}{5}}{6} = \frac{y + \frac{7}{15}}{2} = \frac{z - \frac{3}{10}}{-3}\)
On comparing the above equation with Eq.(ii), we get:
\(a = 6, b = 2\) and \(c = -3\)
which are the direction ratios of the given line.
Now, the direction cosines of given line are:
\(\frac{6}{\sqrt{6^2 + 2^2 + (-3)^2}}\), \(\frac{2}{\sqrt{6^2 + 2^2 + (-3)^2}}\) and \(\frac{-3}{\sqrt{6^2 + 2^2 + (-3)^2}}\) i.e. \(\frac{6}{7}, \frac{2}{7}, \frac{-3}{7}\).

Question. Show that the following points are collinear: \(P(1, -2, -2), Q(3, -6, -8)\) and \(R(-2, 4, 7)\).
Answer: Given points are \(P(1, -2, -2), Q(3, -6, -8)\) and \(R(-2, 4, 7)\).
Direction ratios of line joining \(P\) and \(Q\) are:
\([3 - 1, -6 - (-2), -8 - (-2)]\) i.e. \((2, -4, -6)\).
Direction ratios of line joining \(Q\) and \(R\) are:
\([-2 - 3, 4 + 6, 7 + 8]\) i.e. \((-5, 10, 15)\).
Now, ratios of direction ratios of \(PQ\) and \(QR\) are:
\(\frac{2}{-5}, \frac{-4}{10}, \frac{-6}{15}\) i.e. \(-\frac{2}{5}, -\frac{2}{5}, -\frac{2}{5}\).
Thus, the direction ratios of \(PQ\) and \(QR\) are proportional.
So, \(PQ\) is parallel to \(QR\) but \(Q\) is common to both \(PQ\) and \(QR\).
Hence, \(P, Q\) and \(R\) are collinear.

Question. A line passes through the point with position vector \(2\hat{i} - 3\hat{j} + 4\hat{k}\) and makes angles \(60^\circ\), \(120^\circ\) and \(45^\circ\) with \(X, Y\) and \(Z\)-axes, respectively. Find the equation of the line in the cartesian form.
Answer: The coordinates of the position vector \(2\hat{i} - 3\hat{j} + 4\hat{k}\) is \((2, -3, 4)\).
Also, it is given, \(l = \cos 60^\circ, m = \cos 120^\circ\) and \(n = \cos 45^\circ\)
i.e., \(l = \frac{1}{2}, m = -\frac{1}{2}\) and \(n = \frac{1}{\sqrt{2}}\).
The cartesian equation of line passes through point \((x_1, y_1, z_1)\) and having direction cosines \((l, m, n)\) is:
\(\frac{x-x_1}{l} = \frac{y-y_1}{m} = \frac{z-z_1}{n}\)
Here, \((x_1, y_1, z_1) = (2, -3, 4)\) and \((l, m, n) = \left(\frac{1}{2}, -\frac{1}{2}, \frac{1}{\sqrt{2}}\right)\).
\(\therefore\) The required equation of line is:
\(\frac{x-2}{\frac{1}{2}} = \frac{y+3}{-\frac{1}{2}} = \frac{z-4}{\frac{1}{\sqrt{2}}}\)

Question. Find the vector equation of the line which is parallel to vector \(5\hat{i} + \hat{j} - 7\hat{k}\) and passing through the point \((5, 2, -4)\).
Answer: Equation of a line passing through a point with position vector \(\vec{a}\) and parallel to a vector \(\vec{b}\) is \(\vec{r} = \vec{a} + \lambda\vec{b}\).
Since, line passes through the point \((5, 2, -4)\) and parallel to \(5\hat{i} + \hat{j} - 7\hat{k}\).
\(\therefore \vec{a} = 5\hat{i} + 2\hat{j} - 4\hat{k}\) and \(\vec{b} = 5\hat{i} + \hat{j} - 7\hat{k}\)
\(\therefore\) Equation of line is \(\vec{r} = (5\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(5\hat{i} + \hat{j} - 7\hat{k})\).

Question. Find the equation of a line in cartesian form, which is parallel to vector \(2\hat{i} - \hat{j} + 3\hat{k}\) and passing through a point with position vector \(5\hat{i} - 2\hat{j} + 4\hat{k}\).
Answer: The given line passes through the point having position vector \(\vec{a} = 5\hat{i} - 2\hat{j} + 4\hat{k}\) and is parallel to the vector \(\vec{b} = 2\hat{i} - \hat{j} + 3\hat{k}\).
\(\therefore\) The equation of given line is:
\(\vec{r} = \vec{a} + \lambda\vec{b} \Rightarrow \vec{r} = (5\hat{i} - 2\hat{j} + 4\hat{k}) + \lambda(2\hat{i} - \hat{j} + 3\hat{k})\) ...(i)
For cartesian equation, put \(\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}\) in Eq. (i), we get:
\((x\hat{i} + y\hat{j} + z\hat{k}) = (5\hat{i} - 2\hat{j} + 4\hat{k}) + \lambda(2\hat{i} - \hat{j} + 3\hat{k})\)
\(x\hat{i} + y\hat{j} + z\hat{k} = (5 + 2\lambda)\hat{i} + (-2 - \lambda)\hat{j} + (4 + 3\lambda)\hat{k}\)
On equating the coefficients of \(\hat{i}, \hat{j}\) and \(\hat{k}\) both sides, we get:
\(x = 5 + 2\lambda\), \(y = -2 - \lambda\), \(z = 4 + 3\lambda\)
\(\Rightarrow \frac{x-5}{2} = \lambda\), \(\frac{y+2}{-1} = \lambda\), \(\frac{z-4}{3} = \lambda\)
\(\Rightarrow \frac{x-5}{2} = \frac{y+2}{-1} = \frac{z-4}{3}\), which is the required equation of the given line in cartesian form.

Question. Find the vector equation for the line which passes through the point \((1, 2, 3)\) and is parallel to the line \(\frac{x-1}{-2} = \frac{1-y}{3} = \frac{3-z}{-4}\).
Answer: Given line is passing through the point \((1, 2, 3)\).
\(\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}\)
and parallel to the line \(\frac{x-1}{-2} = \frac{1-y}{3} = \frac{3-z}{-4}\) or \(\frac{x-1}{-2} = \frac{y-1}{-3} = \frac{z-3}{4}\).
Direction ratios of the above line are \((-2, -3, 4)\).
So, required line is parallel to the vector \(\vec{b} = -2\hat{i} - 3\hat{j} + 4\hat{k}\).
\(\because\) Equation of a line passing through the point having position vector \(\vec{a}\) and parallel to \(\vec{b}\) is \(\vec{r} = \vec{a} + \lambda\vec{b}\).
Thus, required equation is:
\(\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(-2\hat{i} - 3\hat{j} + 4\hat{k})\).

Question. If the cartesian equation of a line is \(\frac{3-x}{5} = \frac{y+4}{9} = \frac{2z-6}{4}\), then write the vector equation for the line.
Answer: Given cartesian equation of a line is \(\frac{3-x}{5} = \frac{y+4}{9} = \frac{2z-6}{4}\).
On rewriting the given equation in standard form, we get:
\(\frac{x-3}{-5} = \frac{y+4}{9} = \frac{z-3}{2} = \lambda\) (say)
\(\Rightarrow x = -5\lambda + 3, y = 9\lambda - 4\) and \(z = 2\lambda + 3\)
Now, \(x\hat{i} + y\hat{j} + z\hat{k} = (-5\lambda + 3)\hat{i} + (9\lambda - 4)\hat{j} + (2\lambda + 3)\hat{k}\)
\(\vec{r} = (3\hat{i} - 4\hat{j} + 3\hat{k}) + \lambda(-5\hat{i} + 9\hat{j} + 2\hat{k})\), which is the required equation of a line in vector form.

Question. Find the angle between the lines \(\vec{r} = 2\hat{i} - 5\hat{j} + \hat{k} + \lambda(3\hat{i} + 2\hat{j} + 6\hat{k})\) and \(\vec{r} = 7\hat{i} - 6\hat{k} + \mu(\hat{i} + 2\hat{j} + 2\hat{k})\).
Answer: Given lines are \(\vec{r} = 2\hat{i} - 5\hat{j} + \hat{k} + \lambda(3\hat{i} + 2\hat{j} + 6\hat{k})\) and \(\vec{r} = 7\hat{i} - 6\hat{k} + \mu(\hat{i} + 2\hat{j} + 2\hat{k})\).
On comparing with \(\vec{r} = \vec{a}_1 + \lambda\vec{b}_1\) and \(\vec{r} = \vec{a}_2 + \mu\vec{b}_2\), we get:
\(\vec{b}_1 = 3\hat{i} + 2\hat{j} + 6\hat{k}\) and \(\vec{b}_2 = \hat{i} + 2\hat{j} + 2\hat{k}\).
\(\therefore\) The angle between the lines is given by:
\(\cos\theta = \frac{|\vec{b}_1 \cdot \vec{b}_2|}{|\vec{b}_1||\vec{b}_2|} = \frac{|(3\hat{i} + 2\hat{j} + 6\hat{k}) \cdot (\hat{i} + 2\hat{j} + 2\hat{k})|}{\sqrt{3^2 + 2^2 + 6^2}\sqrt{1^2 + 2^2 + 2^2}}\)
\(= \frac{3+4+12}{\sqrt{9+4+36}\sqrt{1+4+4}} = \frac{19}{\sqrt{49}\sqrt{9}} = \frac{19}{7 \times 3} = \frac{19}{21}\)
\(\therefore \theta = \cos^{-1}\left(\frac{19}{21}\right)\).

Question. Find the angle between the lines \(\frac{x}{2} = \frac{y}{2} = \frac{z}{1}\) and \(\frac{x-5}{4} = \frac{y-2}{1} = \frac{z-3}{8}\).
Answer: Given equations of lines are \(\frac{x}{2} = \frac{y}{2} = \frac{z}{1}\) and \(\frac{x-5}{4} = \frac{y-2}{1} = \frac{z-3}{8}\).
Here, direction ratios of two lines are \((2, 2, 1)\) and \((4, 1, 8)\).
Let \(\theta\) be the acute angle between the given lines, then:
\(\cos\theta = \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2}\sqrt{a_2^2 + b_2^2 + c_2^2}}\)
\(\Rightarrow \cos\theta = \frac{|2\times4 + 2\times1 + 1\times8|}{\sqrt{2^2 + 2^2 + 1^2}\sqrt{4^2 + 1^2 + 8^2}}\)
\(= \frac{|8 + 2 + 8|}{\sqrt{4 + 4 + 1}\sqrt{16 + 1 + 64}} = \frac{18}{\sqrt{9}\sqrt{81}} = \frac{18}{3 \times 9} = \frac{2}{3}\)
\(\therefore \theta = \cos^{-1}\left(\frac{2}{3}\right)\).

Question. Find the value of \(\mu\), so that lines \(\vec{r} = (3\hat{i} - \hat{j} + 2\hat{k}) + \lambda(2\hat{i} - 3\mu\hat{j} + 4\hat{k})\) and \(\vec{r} = -2\hat{i} + 4\hat{j} - 5\hat{k} + \lambda(2\mu\hat{i} + 4\hat{j} + 4\hat{k})\) are perpendicular to each other.
Answer: Given vector equations are
\(\vec{r} = (3\hat{i} - \hat{j} + 2\hat{k}) + \lambda(2\hat{i} - 3\mu\hat{j} + 4\hat{k})\)
and \(\vec{r} = -2\hat{i} + 4\hat{j} - 5\hat{k} + \lambda(2\mu\hat{i} + 4\hat{j} + 4\hat{k})\).
Here, direction ratios of given lines are
\(\vec{b}_1 = 2\hat{i} - 3\mu\hat{j} + 4\hat{k}\) and \(\vec{b}_2 = 2\mu\hat{i} + 4\hat{j} + 4\hat{k}\).
The condition for perpendicular lines is \(\vec{b}_1 \cdot \vec{b}_2 = 0\)
\(\Rightarrow (2\hat{i} - 3\mu\hat{j} + 4\hat{k}) \cdot (2\mu\hat{i} + 4\hat{j} + 4\hat{k}) = 0\)
\(\Rightarrow 4\mu - 12\mu + 16 = 0\)
\(\Rightarrow -8\mu + 16 = 0\)
\(\Rightarrow \mu = 2\)

Question. Find the value of \(k\) so that the lines joining the points \((1, -1, 2)\) and \((3, 4, k)\) is perpendicular to the line joining the points \((0, 3, 2)\) and \((3, 5, 6)\).
Answer: Let \(A(1, -1, 2)\) and \(B(3, 4, k)\).
Then, direction ratios of line \(AB\) are
\((3 - 1, 4 + 1, k - 2)\) i.e. \((2, 5, k - 2)\).
Again let \(C(0, 3, 2)\) and \(D(3, 5, 6)\).
Then, direction ratios of line \(CD\) are
\((3 - 0, 5 - 3, 6 - 2)\) i.e. \((3, 2, 4)\).
\(\because AB \perp CD\)
\(\therefore a_1 a_2 + b_1 b_2 + c_1 c_2 = 0\) [direction ratio form]
\(\Rightarrow 2 \times 3 + 5 \times 2 + 4(k - 2) = 0\)
\(\Rightarrow 6 + 10 + 4k - 8 = 0\)
\(\Rightarrow k = -\frac{8}{4}\)
\(\Rightarrow k = -2\)

Question. Show that the shortest distance between the lines \(\vec{r} = -\hat{i} + 4\hat{j} + 3\hat{k} + \lambda(4\hat{i} + \hat{j})\) and \(\vec{r} = -4\hat{i} + \hat{j} + \mu(5\hat{i} + 2\hat{j} + \hat{k})\) is zero.
Answer: Given equation of lines are
\(\vec{r} = -\hat{i} + 4\hat{j} + 3\hat{k} + \lambda(4\hat{i} + \hat{j})\)
and \(\vec{r} = (-4\hat{i} + \hat{j}) + \mu(5\hat{i} + 2\hat{j} + \hat{k})\).
On comparing with standard equation \(\vec{r} = \vec{a} + \lambda\vec{b}\), we get
\(\vec{a}_1 = -\hat{i} + 4\hat{j} + 3\hat{k}\), \(\vec{b}_1 = 4\hat{i} + \hat{j}\)
and \(\vec{a}_2 = (-4\hat{i} + \hat{j})\), \(\vec{b}_2 = 5\hat{i} + 2\hat{j} + \hat{k}\).
The condition, for the shortest distance between given lines is zero, is
\((\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = 0\) ...(i)
Now, \(\vec{a}_2 - \vec{a}_1 = (-4\hat{i} + \hat{j}) - (-\hat{i} + 4\hat{j} + 3\hat{k}) = -3\hat{i} - 3\hat{j} - 3\hat{k}\)
and \(\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 1 & 0 \\ 5 & 2 & 1 \end{vmatrix} = \hat{i}(1 - 0) - \hat{j}(4 - 0) + \hat{k}(8 - 5) = \hat{i} - 4\hat{j} + 3\hat{k}\).
From Eq. (i), \((-3\hat{i} - 3\hat{j} - 3\hat{k}) \cdot (\hat{i} - 4\hat{j} + 3\hat{k}) = 0\)
\(\Rightarrow -3 + 12 - 9 = 0\)
\(\Rightarrow 0 = 0\), which is true. Hence proved.

Question. Find the shortest distance between the lines \(\vec{r} = (2\hat{i} - \hat{j} + 3\hat{k}) + \lambda(\hat{i} - 2\hat{j} + 3\hat{k})\) and \(\vec{r} = (\hat{i} + 4\hat{k}) + \mu(3\hat{i} - 6\hat{j} + 9\hat{k})\).
Answer: Given lines are
\(\vec{r} = (2\hat{i} - \hat{j} + 3\hat{k}) + \lambda(\hat{i} - 2\hat{j} + 3\hat{k})\) ...(i)
and \(\vec{r} = (\hat{i} + 4\hat{k}) + \mu(3\hat{i} - 6\hat{j} + 9\hat{k}) = (\hat{i} + 4\hat{k}) + 3\mu(\hat{i} - 2\hat{j} + 3\hat{k})\) ...(ii)
On comparing Eqs. (i) and (ii) with \(\vec{r} = \vec{a}_1 + \lambda\vec{b}\) and \(\vec{r} = \vec{a}_2 + \lambda'\vec{b}\), we get
\(\vec{a}_1 = 2\hat{i} - \hat{j} + 3\hat{k}\), \(\vec{b} = \hat{i} - 2\hat{j} + 3\hat{k}\)
\(\vec{a}_2 = \hat{i} + 4\hat{k}\), \(\vec{b} = \hat{i} - 2\hat{j} + 3\hat{k}\).
Here, the lines are parallel.
\(\therefore\) The shortest distance = \(\frac{|(\vec{a}_2 - \vec{a}_1) \times \vec{b}|}{|\vec{b}|}\)
\(= \frac{|(-\hat{i} + \hat{j} + \hat{k}) \times (\hat{i} - 2\hat{j} + 3\hat{k})|}{\sqrt{1 + 4 + 9}}\)
Here, \((-\hat{i} + \hat{j} + \hat{k}) \times (\hat{i} - 2\hat{j} + 3\hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 1 \\ 1 & -2 & 3 \end{vmatrix}\)
\(= \hat{i}(3 + 2) - \hat{j}(-3 - 1) + \hat{k}(2 - 1) = 5\hat{i} + 4\hat{j} + \hat{k}\).
Hence, the required shortest distance = \(\frac{|5\hat{i} + 4\hat{j} + \hat{k}|}{\sqrt{14}} = \frac{\sqrt{25 + 16 + 1}}{\sqrt{14}} = \frac{\sqrt{42}}{\sqrt{14}} = \sqrt{3}\) units

Question. Find the shortest distance between the lines \(\frac{x+1}{2} = \frac{y-1}{1} = \frac{z-9}{-3}\) and \(\frac{x-3}{2} = \frac{y+15}{-7} = \frac{z-9}{5}\).
Answer: The given lines are \(\frac{x+1}{2} = \frac{y-1}{1} = \frac{z-9}{-3}\) and \(\frac{x-3}{2} = \frac{y+15}{-7} = \frac{z-9}{5}\).
\(\dots\) The equations of given lines in vector form are
\(\vec{r} = (-\hat{i} + \hat{j} + 9\hat{k}) + \lambda(2\hat{i} + \hat{j} - 3\hat{k})\)
and \(\vec{r} = (3\hat{i} - 15\hat{j} + 9\hat{k}) + \mu(2\hat{i} - 7\hat{j} + 5\hat{k})\).
Here, \(\vec{a}_1 = -\hat{i} + \hat{j} + 9\hat{k}\), \(\vec{b}_1 = 2\hat{i} + \hat{j} - 3\hat{k}\)
and \(\vec{a}_2 = 3\hat{i} - 15\hat{j} + 9\hat{k}\), \(\vec{b}_2 = 2\hat{i} - 7\hat{j} + 5\hat{k}\).
Now, \(\vec{a}_2 - \vec{a}_1 = 4\hat{i} - 16\hat{j}\)
and \(\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -3 \\ 2 & -7 & 5 \end{vmatrix} = \hat{i}(5 - 21) - \hat{j}(10 + 6) + \hat{k}(-14 - 2) = -16\hat{i} - 16\hat{j} - 16\hat{k}\).
Here, \(|\vec{b}_1 \times \vec{b}_2| = \sqrt{256 + 256 + 256} = 16\sqrt{3}\).
\(\therefore \text{SD} = \frac{|(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1)|}{|\vec{b}_1 \times \vec{b}_2|}\)
\(= \frac{|(-16\hat{i} - 16\hat{j} - 16\hat{k}) \cdot (4\hat{i} - 16\hat{j})|}{16\sqrt{3}} = \frac{|-64 + 256|}{16\sqrt{3}} = \frac{192}{16\sqrt{3}} = \frac{12}{\sqrt{3}} = 4\sqrt{3}\) units

Question. Find the coordinates of the foot of the perpendicular drawn from point \(P(5,7,3)\) to the line \(\frac{x-15}{3} = \frac{y-29}{8} = \frac{z-5}{-5}\).
Answer: Given line is \(\frac{x-15}{3} = \frac{y-29}{8} = \frac{z-5}{-5} = \lambda\) (say)
\(\Rightarrow \frac{x-15}{3} = \lambda\), \(\frac{y-29}{8} = \lambda\) and \(\frac{z-5}{-5} = \lambda\)
\(\Rightarrow x = 3\lambda + 15\), \(y = 8\lambda + 29\) and \(z = -5\lambda + 5\)
\(\therefore\) Coordinate of point \(L\) are \(\{(3\lambda + 15), (8\lambda + 29), (-5\lambda + 5)\}\).
Direction ratios of line \(PL\)
\(= (3\lambda + 15 - 5, 8\lambda + 29 - 7, -5\lambda + 5 - 3) = (3\lambda + 10, 8\lambda + 22, -5\lambda + 2)\).
Direction ratios of line \(AB\) are \((3, 8, -5)\).
\(\because PL \perp AB\)
\(\therefore a_1a_2 + b_1b_2 + c_1c_2 = 0\)
\(\Rightarrow 3(3\lambda + 10) + 8(8\lambda + 22) - 5(-5\lambda + 2) = 0\)
\(\Rightarrow 9\lambda + 30 + 64\lambda + 176 + 25\lambda - 10 = 0\)
\(\Rightarrow 98\lambda + 196 = 0 \Rightarrow \lambda = \frac{-196}{98} = -2\).
\(\therefore\) Foot of perpendicular, \(L = (3\lambda + 15, 8\lambda + 29, -5\lambda + 5) = (9, 13, 15)\)

Long Answer Type Questions

Question. Find the vector and cartesian equations of the line through the point \((1, 2, -4)\) and perpendicular to the two lines \(\vec{r} = (8\hat{i} - 19\hat{j} + 10\hat{k}) + \lambda(3\hat{i} - 16\hat{j} + 7\hat{k})\) and \(\vec{r} = (15\hat{i} + 29\hat{j} + 5\hat{k}) + \mu(3\hat{i} + 8\hat{j} - 5\hat{k})\).
Answer: Given equations of lines are
\(\vec{r} = (8\hat{i} - 19\hat{j} + 10\hat{k}) + \lambda(3\hat{i} - 16\hat{j} + 7\hat{k})\)
and \(\vec{r} = (15\hat{i} + 29\hat{j} + 5\hat{k}) + \mu(3\hat{i} + 8\hat{j} - 5\hat{k})\).
On comparing with \(\vec{r} = \vec{a}_1 + \lambda\vec{b}_1\) and \(\vec{r} = \vec{a}_2 + \mu\vec{b}_2\), we get
\(\vec{b}_1 = 3\hat{i} - 16\hat{j} + 7\hat{k}\) and \(\vec{b}_2 = 3\hat{i} + 8\hat{j} - 5\hat{k}\).
Now, we determine
\(\vec{b} = \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{vmatrix} = \hat{i}(80 - 56) - \hat{j}(-15 - 21) + \hat{k}(24 + 48) = 24\hat{i} + 36\hat{j} + 72\hat{k} = 12(2\hat{i} + 3\hat{j} + 6\hat{k})\).
Since, the required line is perpendicular to the given lines, so it is parallel to \(\vec{b}_1 \times \vec{b}_2\).
Now, equation of a line passing through the point \((1, 2, -4)\) and parallel to \(24\hat{i} + 36\hat{j} + 72\hat{k}\) or \((2\hat{i} + 3\hat{j} + 6\hat{k})\) is
\(\vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})\), which is required vector equation of the line.
For cartesian equation,
On putting \(\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}\), we get
\(x\hat{i} + y\hat{j} + z\hat{k} = (1 + 2\lambda)\hat{i} + (2 + 3\lambda)\hat{j} + (-4 + 6\lambda)\hat{k}\).
On comparing the coefficients of \(\hat{i}\), \(\hat{j}\) and \(\hat{k}\), we get
\(x = 1 + 2\lambda\), \(y = 2 + 3\lambda\) and \(z = -4 + 6\lambda\)
\(\Rightarrow \frac{x-1}{2} = \lambda\), \(\frac{y-2}{3} = \lambda\) and \(\frac{z+4}{6} = \lambda\)
\(\Rightarrow \frac{x-1}{2} = \frac{y-2}{3} = \frac{z+4}{6}\), which is the required cartesian equation of the line.

Question. A line passes through the point \((2, -1, 3)\) and is perpendicular to the lines \(\vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(2\hat{i} - 2\hat{j} + \hat{k})\) and \(\vec{r} = (2\hat{i} - \hat{j} - 3\hat{k}) + \mu(\hat{i} + 2\hat{j} + 2\hat{k})\). Obtain its equation in vector and cartesian form.
Answer: Let the required equation of line passing through the point \((2, -1, 3)\) and having direction ratios \(\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}\) be
\(\vec{r} = (2\hat{i} - \hat{j} + 3\hat{k}) + \lambda\vec{b}\) ...(i)
The direction ratios of given lines are \(\vec{b}_1 = 2\hat{i} - 2\hat{j} + \hat{k}\) and \(\vec{b}_2 = \hat{i} + 2\hat{j} + 2\hat{k}\).
Since, line (i) is perpendicular to both the given lines,
\(\therefore \vec{b} \cdot \vec{b}_1 = 0\) and \(\vec{b} \cdot \vec{b}_2 = 0\)
\(\Rightarrow (b_1\hat{i} + b_2\hat{j} + b_3\hat{k}) \cdot (2\hat{i} - 2\hat{j} + \hat{k}) = 0\) and \((b_1\hat{i} + b_2\hat{j} + b_3\hat{k}) \cdot (\hat{i} + 2\hat{j} + 2\hat{k}) = 0\)
\(\Rightarrow 2b_1 - 2b_2 + b_3 = 0\) ...(ii)
and \(b_1 + 2b_2 + 2b_3 = 0\) ...(iii)
By cross-multiplication method, solving Eqs. (ii) and (iii), we get
\(\frac{b_1}{-4 - 2} = \frac{b_2}{-(4 - 1)} = \frac{b_3}{4 + 2} \Rightarrow \frac{b_1}{-6} = \frac{b_2}{-3} = \frac{b_3}{6} \Rightarrow \frac{b_1}{-2} = \frac{b_2}{-1} = \frac{b_3}{2}\).
\(\therefore \vec{b} = -2\hat{i} - \hat{j} + 2\hat{k}\).
On putting the value of \(\vec{b}\) in Eq. (i), we get \(\vec{r} = 2\hat{i} - \hat{j} + 3\hat{k} + \lambda(-2\hat{i} - \hat{j} + 2\hat{k})\).
The cartesian equation of a line passing through \((2, -1, 3)\) and having direction ratios \((-2, -1, 2)\) is \(\frac{x-2}{-2} = \frac{y+1}{-1} = \frac{z-3}{2}\).

Question. Find the vector and cartesian equations of the line which is perpendicular to the lines with equations \(\frac{x+2}{1} = \frac{y-3}{2} = \frac{z+1}{4}\) and \(\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}\) and passes through the point \((1, 1, 1)\). Also, find the angle between the given lines.
Answer: Any line through the point \((1, 1, 1)\) is given by
\(\frac{x-1}{a} = \frac{y-1}{b} = \frac{z-1}{c}\) ...(i)
Where \(a, b\) and \(c\) are the direction ratios of line (i).
Now, the line (i) is perpendicular to the lines \(\frac{x+2}{1} = \frac{y-3}{2} = \frac{z+1}{4}\) and \(\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}\) where direction ratios of these two lines are \((1, 2, 4)\) and \((2, 3, 4)\), respectively.
\(\therefore a + 2b + 4c = 0\) ...(ii)
and \(2a + 3b + 4c = 0\) ...(iii)
[\(\because\) if two lines having direction ratios \((a_1, b_1, c_1)\) and \((a_2, b_2, c_2)\) are perpendicular, then \(a_1a_2 + b_1b_2 + c_1c_2 = 0\)]
By cross-multiplication method, we get
\(\frac{a}{8 - 12} = \frac{b}{-(4 - 8)} = \frac{c}{3 - 4} \Rightarrow \frac{a}{-4} = \frac{b}{4} = \frac{c}{-1}\).
\(\therefore\) Direction ratios of line (i) are \(-4, 4, -1\).
\(\therefore\) The required cartesian equation of line (i) is \(\frac{x-1}{-4} = \frac{y-1}{4} = \frac{z-1}{-1}\) and vector equation is \(\vec{r} = \hat{i} + \hat{j} + \hat{k} + \lambda(-4\hat{i} + 4\hat{j} - \hat{k})\).
Again, let \(\theta\) be the angle between the given lines.
\(\therefore \cos \theta = \frac{|1 \times 2 + 2 \times 3 + 4 \times 4|}{\sqrt{1 + 4 + 16}\sqrt{4 + 9 + 16}} = \frac{24}{\sqrt{21}\sqrt{29}} = \frac{24}{\sqrt{609}}\)
\(\theta = \cos^{-1}\left(\frac{24}{\sqrt{609}}\right)\)

Question. Show that the lines \(\frac{x+1}{3} = \frac{y+3}{5} = \frac{z+5}{7}\) and \(\frac{x-2}{1} = \frac{y-4}{3} = \frac{z-6}{5}\) intersect. Also, find their points of intersection.
Answer: The given lines are \(\frac{x+1}{3} = \frac{y+3}{5} = \frac{z+5}{7} = \lambda\) (let) ...(i)
and \(\frac{x-2}{1} = \frac{y-4}{3} = \frac{z-6}{5} = \mu\) (let) ...(ii)
Then, any point on line (i) is \(P(3\lambda - 1, 5\lambda - 3, 7\lambda - 5)\) ...(iii)
and any point on line (ii) is \(Q(\mu + 2, 3\mu + 4, 5\mu + 6)\) ...(iv)
Clearly, the lines (i) and (ii) will intersect, if \((3\lambda - 1, 5\lambda - 3, 7\lambda - 5) = (\mu + 2, 3\mu + 4, 5\mu + 6)\) for some particular value of \(\lambda\) and \(\mu\).
\(\Rightarrow 3\lambda - 1 = \mu + 2 \Rightarrow 3\lambda - \mu = 3\) ...(v)
\(5\lambda - 3 = 3\mu + 4 \Rightarrow 5\lambda - 3\mu = 7\) ...(vi)
and \(7\lambda - 5 = 5\mu + 6 \Rightarrow 7\lambda - 5\mu = 11\) ...(vii)
On multiplying Eq. (v) by 3 and then subtracting Eq. (vi) from it, we get
\(9\lambda - 3\mu - 5\lambda + 3\mu = 9 - 7 \Rightarrow 4\lambda = 2 \Rightarrow \lambda = \frac{1}{2}\).
On putting the value of \(\lambda\) in Eq. (v), we get \(3 \times \frac{1}{2} - \mu = 3 \Rightarrow \frac{3}{2} - \mu = 3 \Rightarrow \mu = -\frac{3}{2}\).
On putting the values of \(\lambda\) and \(\mu\) in Eq. (vii), we get \(7 \times \frac{1}{2} - 5\left(-\frac{3}{2}\right) = 11 \Rightarrow \frac{7}{2} + \frac{15}{2} = 11 \Rightarrow \frac{22}{2} = 11 \Rightarrow 11 = 11\), which is true.
Hence, lines (i) and (ii) intersect and their point of intersection is \(P\left(3 \times \frac{1}{2} - 1, 5 \times \frac{1}{2} - 3, 7 \times \frac{1}{2} - 5\right)\) i.e. \(P\left(\frac{1}{2}, -\frac{1}{2}, -\frac{3}{2}\right)\).

Question. Find the value of \(\lambda\) so that the lines \(\frac{1-x}{3} = \frac{7y-14}{\lambda} = \frac{z-3}{2}\) and \(\frac{7-7x}{3\lambda} = \frac{y-5}{1} = \frac{6-z}{5}\) are at right angles. Also, find whether the lines are intersecting or not.
Answer: Given equations of lines can be written in standard form as
\(\frac{x-1}{-3} = \frac{y-2}{\lambda/7} = \frac{z-3}{2} = r_1\) (let) ...(i)
and \(\frac{x-1}{-3\lambda/7} = \frac{y-5}{1} = \frac{z-6}{-5} = r_2\) (let) ...(ii)
These lines will intersect at right angle, if \(-3\left(\frac{-3\lambda}{7}\right) + \left(\frac{\lambda}{7}\right)(1) + 2(-5) = 0\)
[\(\because\) two lines with direction ratios \(a_1,b_1,c_1\) and \(a_2,b_2,c_2\) are perpendicular, if \(a_1a_2 + b_1b_2 + c_1c_2 = 0\)]
\(\Rightarrow \frac{9\lambda}{7} + \frac{\lambda}{7} = 10 \Rightarrow \frac{10\lambda}{7} = 10 \Rightarrow \lambda = 7\), which is the required value of \(\lambda\).
Now, let us check whether the lines are intersecting or not.
Coordinates of any point on line (i) are \((-3r_1 + 1, r_1 + 2, 2r_1 + 3)\) and coordinates of any point on line (ii) are \((-3r_2 + 1, r_2 + 5, -5r_2 + 6)\).
Clearly, the line will intersect if \((-3r_1 + 1, r_1 + 2, 2r_1 + 3) = (-3r_2 + 1, r_2 + 5, -5r_2 + 6)\) for some \(r_1, r_2 \in \mathbb{R}\).
\(\Rightarrow -3r_1 + 1 = -3r_2 + 1\), \(r_1 + 2 = r_2 + 5\) and \(2r_1 + 3 = -5r_2 + 6\)
\(\Rightarrow r_1 = r_2\), \(r_1 - r_2 = 3\) and \(2r_1 + 5r_2 = 3\), which is not possible simultaneously for any \(r_1, r_2 \in \mathbb{R}\). Hence, the lines are not intersecting.

Question. Find the shortest distance between the lines \(\frac{x-8}{3} = \frac{y+9}{-16} = \frac{z-10}{7}\) and \(\frac{x-15}{3} = \frac{y-29}{8} = \frac{z-5}{-5}\).
Answer: The given lines are \(\frac{x-8}{3} = \frac{y+9}{-16} = \frac{z-10}{7}\) and \(\frac{x-15}{3} = \frac{y-29}{8} = \frac{z-5}{-5}\).
\(\therefore\) The equation of given lines in vector form are \(\vec{r} = 8\hat{i} - 9\hat{j} + 10\hat{k} + \lambda(3\hat{i} - 16\hat{j} + 7\hat{k})\) and \(\vec{r} = 15\hat{i} + 29\hat{j} + 5\hat{k} + \mu(3\hat{i} + 8\hat{j} - 5\hat{k})\).
Here, \(\vec{a}_1 = 8\hat{i} - 9\hat{j} + 10\hat{k}\), \(\vec{b}_1 = 3\hat{i} - 16\hat{j} + 7\hat{k}\), \(\vec{a}_2 = 15\hat{i} + 29\hat{j} + 5\hat{k}\) and \(\vec{b}_2 = 3\hat{i} + 8\hat{j} - 5\hat{k}\).
Now, \(\vec{a}_2 - \vec{a}_1 = 7\hat{i} + 38\hat{j} - 5\hat{k}\) and \(\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{vmatrix} = \hat{i}(80 - 56) - \hat{j}(-15 - 21) + \hat{k}(24 + 48) = 24\hat{i} + 36\hat{j} + 72\hat{k}\).
Here, \(|\vec{b}_1 \times \vec{b}_2| = \sqrt{576 + 1296 + 5184} = \sqrt{7056} = 84\).
\(\therefore \text{SD} = \frac{|(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1)|}{|\vec{b}_1 \times \vec{b}_2|} = \frac{|(24\hat{i} + 36\hat{j} + 72\hat{k}) \cdot (7\hat{i} + 38\hat{j} - 5\hat{k})|}{84} = \frac{|168 + 1368 - 360|}{84} = \frac{1176}{84} = 14\) units

Question. Find the shortest distance between the lines \(L_1\) and \(L_2\) given below.
\(L_1\) : The line passing through \((2, -1, 1)\) and parallel to \(\frac{x}{1} = \frac{y}{1} = \frac{z}{3}\).
\(L_2\) : \(\vec{r} = \hat{i} + (2\mu + 1)\hat{j} - (\mu + 2)\hat{k}\).
Answer: Given, \(L_1\) : The line passing through \((2, -1, 1)\) and parallel to \(\frac{x}{1} = \frac{y}{1} = \frac{z}{3}\).
Write the given equations of line in standard form:
\(L_1 : \vec{r} = (2\hat{i} - \hat{j} + \hat{k}) + \lambda(\hat{i} + \hat{j} + 3\hat{k})\) ...(i)
\(L_2 : \vec{r} = (\hat{i} + \hat{j} - 2\hat{k}) + \mu(2\hat{j} - \hat{k})\) ...(ii)
On comparing Eqs. (i) and (ii) with \(\vec{r} = \vec{a}_1 + \lambda\vec{b}_1\) and \(\vec{r} = \vec{a}_2 + \mu\vec{b}_2\) respectively, we get:
\(\vec{a}_1 = 2\hat{i} - \hat{j} + \hat{k}\), \(\vec{b}_1 = \hat{i} + \hat{j} + 3\hat{k}\) and \(\vec{a}_2 = (\hat{i} + \hat{j} - 2\hat{k})\), \(\vec{b}_2 = 2\hat{j} - \hat{k}\).
Clearly, \(\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 3 \\ 0 & 2 & -1 \end{vmatrix} = \hat{i}(-1 - 6) - \hat{j}(-1 - 0) + \hat{k}(2 - 0) = -7\hat{i} + \hat{j} + 2\hat{k}\).
\(\Rightarrow |\vec{b}_1 \times \vec{b}_2| = |-7\hat{i} + \hat{j} + 2\hat{k}| = \sqrt{(-7)^2 + (1)^2 + (2)^2} = \sqrt{49 + 1 + 4} = \sqrt{54} = 3\sqrt{6}\).
Now, \(\vec{a}_2 - \vec{a}_1 = (\hat{i} + \hat{j} - 2\hat{k}) - (2\hat{i} - \hat{j} + \hat{k}) = -\hat{i} + 2\hat{j} - 3\hat{k}\).
\(\therefore \text{Required SD} = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|} = \frac{|(-\hat{i} + 2\hat{j} - 3\hat{k}) \cdot (-7\hat{i} + \hat{j} + 2\hat{k})|}{3\sqrt{6}} = \frac{|7 + 2 - 6|}{3\sqrt{6}} = \frac{3}{3\sqrt{6}} = \frac{1}{\sqrt{6}}\) unit

Question. Vertices B and C of \(\Delta ABC\) lie on the line \(\frac{x+2}{2} = \frac{y-1}{1} = \frac{z}{4}\). Find the area of \(\Delta ABC\) given that point A has coordinates \((1, -1, 2)\) and the line segment BC has length of 5 units.
Answer: Let \(h\) be the height of \(\Delta ABC\). Then, \(h\) is the length of perpendicular from \(A(1, -1, 2)\) to the line \(\frac{x+2}{2} = \frac{y-1}{1} = \frac{z-0}{4}\).
Clearly, the line \(\frac{x+2}{2} = \frac{y-1}{1} = \frac{z-0}{4}\) passes through the point say \(P(-2, 1, 0)\) and parallel to the vector \(\vec{b} = 2\hat{i} + \hat{j} + 4\hat{k}\).
Let \(\frac{x+2}{2} = \frac{y-1}{1} = \frac{z-0}{4} = \lambda\).
Then, coordinates of \(M\) are \((2\lambda - 2, \lambda + 1, 4\lambda)\).
Now, direction ratios of \(AM = (2\lambda - 2 - 1, \lambda + 1 - (-1), 4\lambda - 2) = (2\lambda - 3, \lambda + 2, 4\lambda - 2)\).
Since, \(AM \perp BC\),
\(\therefore 2(2\lambda - 3) + 1(\lambda + 2) + 4(4\lambda - 2) = 0 \Rightarrow 4\lambda - 6 + \lambda + 2 + 16\lambda - 8 = 0 \Rightarrow 21\lambda = 12 \Rightarrow \lambda = \frac{4}{7}\).
Thus, the coordinates of \(M\) are \(\left(2\times\frac{4}{7} - 2, \frac{4}{7} + 1, 4\times\frac{4}{7}\right)\) i.e. \(\left(-\frac{6}{7}, \frac{11}{7}, \frac{16}{7}\right)\).
Now, \(h = |AM| = \sqrt{\left(-\frac{6}{7} - 1\right)^2 + \left(\frac{11}{7} + 1\right)^2 + \left(\frac{16}{7} - 2\right)^2} = \sqrt{\left(-\frac{13}{7}\right)^2 + \left(\frac{18}{7}\right)^2 + \left(\frac{2}{7}\right)^2} = \sqrt{\frac{169}{7^2} + \frac{324}{7^2} + \frac{4}{7^2}} = \sqrt{\frac{497}{7^2}} = \frac{\sqrt{71}}{\sqrt{7}}\).
It is given that the length of \(BC\) is 5 units.
\(\therefore \text{Area of } \Delta ABC = \frac{1}{2} \times (BC \times h) = \frac{1}{2} \times 5 \times \frac{\sqrt{71}}{\sqrt{7}} = \sqrt{\frac{1775}{28}}\) sq units

Question. Find the image of the point \((2, -1, 5)\) in the line \(\frac{x-11}{10} = \frac{y+2}{-4} = \frac{z+8}{-11}\).
Answer: Let \(T\) be the image of the point \(P(2, -1, 5)\). \(Q\) is the foot of perpendicular drawn from point \(P\) on the line \(AB\).
Given equation of line \(AB\) is \(\frac{x-11}{10} = \frac{y+2}{-4} = \frac{z+8}{-11}\) ...(i)
Let \(\frac{x-11}{10} = \frac{y+2}{-4} = \frac{z+8}{-11} = \lambda\) (say) \(\Rightarrow x = 10\lambda + 11, y = -4\lambda - 2\) and \(z = -11\lambda - 8\).
Then, coordinates of \(Q\) are \((10\lambda + 11, -4\lambda - 2, -11\lambda - 8)\) ...(ii)
Now, direction ratios of line \(PQ = (10\lambda + 11 - 2, -4\lambda - 2 + 1, -11\lambda - 8 - 5) = (10\lambda + 9, -4\lambda - 1, -11\lambda - 13)\).
Since, line \(PQ \perp AB\), \(\therefore a_1a_2 + b_1b_2 + c_1c_2 = 0\) where \(a_1 = 10\lambda + 9\), \(b_1 = -4\lambda - 1\), \(c_1 = -11\lambda - 13\) and \(a_2 = 10\), \(b_2 = -4\), \(c_2 = -11\).
\(\therefore (10\lambda + 9)(10) + (-4\lambda - 1)(-4) + (-11\lambda - 13)(-11) = 0 \Rightarrow 100\lambda + 90 + 16\lambda + 4 + 121\lambda + 143 = 0 \Rightarrow 237\lambda + 237 = 0 \Rightarrow \lambda = -1\).
On putting \(\lambda = -1\) in Eq. (ii), we get \(Q = (10(-1) + 11, -4(-1) - 2, -11(-1) - 8) = (1, 2, 3)\).
Let image of point \(P\) be \(T(x, y, z)\). Then, \(Q\) will be the mid-point of \(PT\).
By mid-point formula, \(Q = \text{Mid-point of } P(2, -1, 5) \text{ and } T(x, y, z)\).
\(\Rightarrow \left(\frac{x+2}{2}, \frac{y-1}{2}, \frac{z+5}{2}\right) = (1, 2, 3)\)
On equating corresponding coordinates, we get:
\(\frac{x+2}{2} = 1 \Rightarrow x = 0\), \(\frac{y-1}{2} = 2 \Rightarrow y = 5\), \(\frac{z+5}{2} = 3 \Rightarrow z = 1\).
\(\therefore\) Coordinates of \(T = (x, y, z) = (0, 5, 1)\). Hence, the image of point \(P(2, -1, 5)\) is \(T(0, 5, 1)\).

Question. Find the foot of the perpendicular drawn from the point \(\hat{i} + 6\hat{j} + 3\hat{k}\) to the line \(\vec{r} = \hat{j} + 2\hat{k} + \lambda(\hat{i} + 2\hat{j} + 3\hat{k})\). Also, find the length of the perpendicular.
Answer: Let \(L\) be the foot of the perpendicular drawn from \(P(\hat{i} + 6\hat{j} + 3\hat{k})\) on the line \(\vec{r} = \hat{j} + 2\hat{k} + \lambda(\hat{i} + 2\hat{j} + 3\hat{k})\) ...(i).
Then, \(\vec{PL} = \text{Position vector of } L - \text{Position vector of } P = \lambda\hat{i} + (1 + 2\lambda)\hat{j} + (2 + 3\lambda)\hat{k} - (\hat{i} + 6\hat{j} + 3\hat{k}) = (\lambda - 1)\hat{i} + (-5 + 2\lambda)\hat{j} + (-1 + 3\lambda)\hat{k}\).
Since, \(\vec{PL}\) is perpendicular to the given line which is parallel to \(\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}\), \(\therefore \vec{PL} \perp \vec{b}\).
\(\Rightarrow [(\lambda - 1)\hat{i} + (-5 + 2\lambda)\hat{j} + (-1 + 3\lambda)\hat{k}] \cdot [\hat{i} + 2\hat{j} + 3\hat{k}] = 0\)
\(\Rightarrow (\lambda - 1) \times 1 + (-5 + 2\lambda) \times 2 + (-1 + 3\lambda) \times 3 = 0 \Rightarrow \lambda - 1 - 10 + 4\lambda - 3 + 9\lambda = 0 \Rightarrow 14\lambda - 14 = 0 \Rightarrow \lambda = 1\).
On putting \(\lambda = 1\) in Eq. (i), we get the position vector of \(L\) is \(\hat{j} + 2\hat{k} + 1(\hat{i} + 2\hat{j} + 3\hat{k})\) or \((\hat{i} + 3\hat{j} + 5\hat{k})\).
Now, \(\vec{PL} = (\hat{i} + 3\hat{j} + 5\hat{k}) - (\hat{i} + 6\hat{j} + 3\hat{k}) = -3\hat{j} + 2\hat{k}\).
\(\therefore\) The length of perpendicular drawn from \(P\) to the line \(= |\vec{PL}| = \sqrt{(-3)^2 + (2)^2} = \sqrt{9 + 4} = \sqrt{13}\).

Question. Find the image of the point \((-1, 5, 2)\) in the line \(\frac{2x-4}{2} = \frac{y}{2} = \frac{2-z}{3}\). Find the length of the line segment joining the points (given point and the image).
Answer: Let \(T\) be the image of the point \(P(-1, 5, 2)\). \(Q\) is the foot of perpendicular drawn from point \(P\) on the line \(AB\).
Given equation of line \(AB\) is \(\frac{2x-4}{2} = \frac{y}{2} = \frac{2-z}{3} \Rightarrow \frac{x-2}{1} = \frac{y}{2} = \frac{z-2}{-3}\) ...(i).
Let \(\frac{x-2}{1} = \frac{y}{2} = \frac{z-2}{-3} = \lambda \Rightarrow x = \lambda + 2, y = 2\lambda, z = -3\lambda + 2\).
Then, \(Q = (\lambda + 2, 2\lambda, -3\lambda + 2)\) ...(ii).
Now, direction ratios of line \(PQ = (\lambda + 2 + 1, 2\lambda - 5, -3\lambda + 2 - 2) = (\lambda + 3, 2\lambda - 5, -3\lambda)\).
Since, line \(PQ \perp AB\), \(\therefore a_1a_2 + b_1b_2 + c_1c_2 = 0\), where \(a_1 = \lambda + 3\), \(b_1 = 2\lambda - 5\), \(c_1 = -3\lambda\) and \(a_2 = 1\), \(b_2 = 2\), \(c_2 = -3\).
\(\therefore 1(\lambda + 3) + 2(2\lambda - 5) - 3(-3\lambda) = 0 \Rightarrow \lambda + 3 + 4\lambda - 10 + 9\lambda = 0 \Rightarrow 14\lambda - 7 = 0 \Rightarrow \lambda = \frac{1}{2}\).
On putting \(\lambda = \frac{1}{2}\) in Eq. (ii), we get \(Q = \left(\frac{1}{2} + 2, 2 \times \frac{1}{2}, -3 \times \frac{1}{2} + 2\right) = \left(\frac{5}{2}, 1, \frac{1}{2}\right)\).
Let image of point \(P\) be \(T(x, y, z)\). Then, \(Q\left(\frac{5}{2}, 1, \frac{1}{2}\right)\) will be the mid-point of \(P(-1, 5, 2)\) and \(T(x, y, z)\).
\(\therefore \left(\frac{5}{2}, 1, \frac{1}{2}\right) = \left(\frac{x-1}{2}, \frac{y+5}{2}, \frac{z+2}{2}\right)\).
On equating corresponding coordinates, we get:
\(\frac{x-1}{2} = \frac{5}{2} \Rightarrow x = 6\), \(\frac{y+5}{2} = 1 \Rightarrow y = -3\), \(\frac{z+2}{2} = \frac{1}{2} \Rightarrow z = -1\).
\(\therefore\) Coordinates of \(T = (x, y, z) = (6, -3, -1)\). Hence, the image of point \(P(-1, 5, 2)\) is \(T(6, -3, -1)\).
Now, distance of \(PT = \sqrt{(6 - (-1))^2 + (-3 - 5)^2 + (-1 - 2)^2} = \sqrt{7^2 + (-8)^2 + (-3)^2} = \sqrt{49 + 64 + 9} = \sqrt{122}\).
So, the length of line segment joining the points is \(\sqrt{122}\) units.