CBSE Class 12 Mathematics Three Dimensional Geometry Assignment Set 06

Very Short Answer Type Questions

Question. Find the direction cosines of the line \(\frac{4-x}{2} = \frac{y}{6} = \frac{1-z}{3}\).
Answer: The equation of the given line is \(\frac{4-x}{2} = \frac{y}{6} = \frac{1-z}{3}\) ... (i)
\(\Rightarrow \frac{x-4}{-2} = \frac{y}{6} = \frac{z-1}{-3}\)
As \(\sqrt{(-2)^2 + 6^2 + (-3)^2} = 7\)
\(\therefore\) Direction cosines of (i) are \(-\frac{2}{7}, \frac{6}{7}, -\frac{3}{7}\).

Question. Find the vector equation of a line which passes through the points \((3, 4, -7)\) and \((1, -1, 6)\).
Answer: The vector equation of a line passing through the points \((3, 4, -7)\) and \((1, -1, 6)\) is given by:
\(\vec{r} = (3\hat{i} + 4\hat{j} - 7\hat{k}) + \lambda [(\hat{i} - \hat{j} + 6\hat{k}) - (3\hat{i} + 4\hat{j} - 7\hat{k})]\)
\(\Rightarrow \vec{r} = 3\hat{i} + 4\hat{j} - 7\hat{k} + \lambda (-2\hat{i} - 5\hat{j} + 13\hat{k})\).

Question. The equation of a line are \(5x - 3 = 15y + 7 = 3 - 10z\). Write the direction cosines of the line.
Answer: The given line is \(5x - 3 = 15y + 7 = 3 - 10z\)
\(\Rightarrow \frac{x - \frac{3}{5}}{\frac{1}{5}} = \frac{y + \frac{7}{15}}{\frac{1}{15}} = \frac{z - \frac{3}{10}}{-\frac{1}{10}}\)
Its direction ratios are \(\frac{1}{5}, \frac{1}{15}, -\frac{1}{10}\), which are proportional to \(6, 2, -3\).
Now, \(\sqrt{6^2 + 2^2 + (-3)^2} = 7\).
\(\therefore\) Its direction cosines are \(\frac{6}{7}, \frac{2}{7}, -\frac{3}{7}\).

Question. Find the length of the intercept, cut off by the plane \(2x + y - z = 5\) on the x-axis.
Answer: We have, \(2x + y - z = 5\)
\(\Rightarrow \frac{x}{5/2} + \frac{y}{5} + \frac{z}{-5} = 1\), which is the intercept form of a plane.
\(\therefore\) Intercepts on the x, y, and z-axes respectively are \(\frac{5}{2}, 5, -5\).
\(\therefore\) Required length of intercept on the x-axis = \(\frac{5}{2}\).

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: The vector equation of the line passing through \((1, -1, 2)\) and parallel to the line \(\frac{x-3}{1} = \frac{y-1}{2} = \frac{z+1}{-2}\) is:
\(\vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \lambda (\hat{i} + 2\hat{j} - 2\hat{k})\).

Question. Find the vector equation of a plane which is at a distance of 5 units from the origin and its normal vector is \(2\hat{i} - 3\hat{j} + 6\hat{k}\).
Answer: Let \(\vec{n} = 2\hat{i} - 3\hat{j} + 6\hat{k}\).
\(\therefore \hat{n} = \frac{\vec{n}}{|\vec{n}|} = \frac{2\hat{i} - 3\hat{j} + 6\hat{k}}{\sqrt{4+9+36}} = \frac{2\hat{i} - 3\hat{j} + 6\hat{k}}{7}\).
So, the required equation of the plane is:
\(\vec{r} \cdot \left(\frac{2}{7}\hat{i} - \frac{3}{7}\hat{j} + \frac{6}{7}\hat{k}\right) = 5 \Rightarrow \vec{r} \cdot (2\hat{i} - 3\hat{j} + 6\hat{k}) = 35\).

Question. Write the equation of a plane which is at a distance of \(5\sqrt{3}\) units from origin and the normal to which is equally inclined to coordinate axes.
Answer: Let \(\alpha, \beta\) and \(\gamma\) be the angles made by \(\vec{n}\) with the x, y, and z-axes, respectively.
Given \(\alpha = \beta = \gamma \Rightarrow \cos\alpha = \cos\beta = \cos\gamma\)
\(\Rightarrow l = m = n\), where \(l, m, n\) are direction cosines of \(\vec{n}\).
Since \(l^2 + m^2 + n^2 = 1 \Rightarrow 3l^2 = 1 \Rightarrow l = \pm\frac{1}{\sqrt{3}}\).
So, \(l = m = n = \pm\frac{1}{\sqrt{3}}\).
The normal form of the plane is \(lx + my + nz = d\)
\(\Rightarrow \pm\frac{1}{\sqrt{3}}x \pm\frac{1}{\sqrt{3}}y \pm\frac{1}{\sqrt{3}}z = 5\sqrt{3} \Rightarrow \pm x \pm y \pm z = 15\).

Question. Find the distance between the planes \(2x - y + 2z = 5\) and \(5x - 2.5y + 5z = 20\).
Answer: The equations of the planes can be written as \(2x - y + 2z = 5\) and \(2x - y + 2z = 8\).
Distance between both the planes is:
\(\frac{|8 - 5|}{\sqrt{2^2 + (-1)^2 + 2^2}} = \frac{3}{\sqrt{9}} = 1\) unit.

Question. Find the distance of the plane \(3x - 4y + 12z = 3\) from the origin.
Answer: Perpendicular distance from the origin \((0, 0, 0)\) to the plane \(3x - 4y + 12z - 3 = 0\) is:
\(\frac{|3(0) - 4(0) + 12(0) - 3|}{\sqrt{3^2 + (-4)^2 + 12^2}} = \frac{3}{13}\) unit.

Question. Write the vector equation of the line passing through \((1, 2, 3)\) and perpendicular to the plane \(\vec{r} \cdot (\hat{i} + 2\hat{j} - 5\hat{k}) + 9 = 0\).
Answer: The given plane is \(\vec{r} \cdot (\hat{i} + 2\hat{j} - 5\hat{k}) + 9 = 0 \Rightarrow \vec{n} = \hat{i} + 2\hat{j} - 5\hat{k}\).
Since the line is perpendicular to the plane, its direction ratios are proportional to the normal vector coefficients: \(1, 2, -5\).
Therefore, the equation of the line through \((1, 2, 3)\) is:
\(\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} + 2\hat{j} - 5\hat{k})\).

Short Answer Type Questions - I

Question. The x-coordinate of a point on the line joining the points \(P(2, 2, 1)\) and \(Q(5, 1, -2)\) is 4. Find its z-coordinate.
Answer: Let point \(R\) on the line \(PQ\) divide it in the ratio \(k : 1\).
By section formula, the x-coordinate is:
\(4 = \frac{5k + 2}{k + 1} \Rightarrow 4k + 4 = 5k + 2 \Rightarrow k = 2\).
Now, the z-coordinate of point \(R\) is:
\(z = \frac{-2k + 1}{k + 1} = \frac{-2(2) + 1}{2 + 1} = -1\).
Thus, the z-coordinate of the point is \(-1\).

Question. Find the value of \(k\) so that the lines \(x = -y = kz\) and \(x - 2 = 2y + 1 = -z + 1\) are perpendicular to each other.
Answer: Rewriting the lines in standard form:
\(l_1 : \frac{x-0}{1} = \frac{y-0}{-1} = \frac{z-0}{1/k}\)
\(l_2 : \frac{x-2}{1} = \frac{y + 1/2}{1/2} = \frac{z-1}{-1}\)
Since \(l_1\) is perpendicular to \(l_2\):
\(1(1) + (-1)\left(\frac{1}{2}\right) + \left(\frac{1}{k}\right)(-1) = 0\)
\(\Rightarrow 1 - \frac{1}{2} - \frac{1}{k} = 0 \Rightarrow \frac{1}{2} = \frac{1}{k} \Rightarrow k = 2\).

Question. Find the vector equation of the line passing through the point \(A(1, 2, -1)\) and parallel to the line \(5x - 25 = 14 - 7y = 35z\).
Answer: Standard form of the given line is:
\(\frac{5(x-5)}{1} = \frac{-7(y-2)}{1} = \frac{35z}{1} \Rightarrow \frac{x-5}{1/5} = \frac{y-2}{-1/7} = \frac{z}{1/35}\)
Multiplying the denominators by 35 gives the direction ratios: \(7, -5, 1\).
Therefore, the vector equation of the line through \(A(1, 2, -1)\) is:
\(\vec{r} = (\hat{i} + 2\hat{j} - \hat{k}) + \lambda(7\hat{i} - 5\hat{j} + \hat{k})\).

Question. Find the coordinates of the point where the line through \((-1, 1, -8)\) and \((5, -2, 10)\) crosses the ZX-plane.
Answer: The equation of the line through \(A(-1, 1, -8)\) and \(B(5, -2, 10)\) is:
\(\frac{x+1}{6} = \frac{y-1}{-3} = \frac{z+8}{18} = k\) ... (i)
Any general point on this line is \((6k - 1, -3k + 1, 18k - 8)\).
For the point crossing the ZX-plane, the y-coordinate must be 0:
\(-3k + 1 = 0 \Rightarrow k = \frac{1}{3}\).
Substituting \(k = \frac{1}{3}\) into the general point yields:
\(\left(6\left(\frac{1}{3}\right) - 1, 0, 18\left(\frac{1}{3}\right) - 8\right) = (1, 0, -2)\).

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 point of intersection.
Answer: Any point on the first line is \((3r - 1, 5r - 3, 7r - 5)\) ... (i)
Any point on the second line is \((k + 2, 3k + 4, 5k + 6)\) ... (ii)
For intersection, we equate the coordinates:
\(3r - 1 = k + 2 \Rightarrow 3r - k = 3\)
\(5r - 3 = 3k + 4 \Rightarrow 5r - 3k = 7\)
Solving these equations, we find \(r = \frac{1}{2}\) and \(k = -\frac{3}{2}\).
Substituting these values into the z-coordinate relation:
\(7\left(\frac{1}{2}\right) - 5 = -\frac{3}{2}\) and \(5\left(-\frac{3}{2}\right) + 6 = -\frac{3}{2}\). Since they are equal, the lines intersect.
The point of intersection is \(\left(\frac{1}{2}, -\frac{1}{2}, -\frac{3}{2}\right)\).

Question. Find the perpendicular distance of the point \((1, 0, 0)\) from the line \(\frac{x-1}{2} = \frac{y+1}{-3} = \frac{z+10}{8}\). Also find the coordinates of the foot of the perpendicular and the equation of the perpendicular.
Answer: Let a general point \(R\) on the line be \((2k+1, -3k-1, 8k-10)\) ... (i).
If \(R\) is the foot of the perpendicular from \(P(1, 0, 0)\), then the direction ratios of \(PR\) are \((2k, -3k-1, 8k-10)\).
Since \(PR\) is perpendicular to the line:
\(2(2k) - 3(-3k-1) + 8(8k-10) = 0 \Rightarrow 77k = 77 \Rightarrow k = 1\).
Thus, the foot of the perpendicular is \(R(3, -4, -2)\).
Perpendicular distance \(PR = \sqrt{(3-1)^2 + (-4-0)^2 + (-2-0)^2} = \sqrt{4 + 16 + 4} = 2\sqrt{6}\) units.
The equation of the perpendicular is \(\frac{x-1}{2} = \frac{y}{-4} = \frac{z}{-2}\).

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: Rewriting the lines in standard form:
\(l_1 : \frac{x-1}{-3} = \frac{y-2}{\lambda/7} = \frac{z-3}{2}\)
\(l_2 : \frac{x-1}{-3\lambda/7} = \frac{y-5}{1} = \frac{z-6}{-5}\)
Since the lines are perpendicular:
\((-3)\left(-\frac{3\lambda}{7}\right) + \left(\frac{\lambda}{7}\right)(1) + 2(-5) = 0 \Rightarrow \frac{10\lambda}{7} = 10 \Rightarrow \lambda = 7\).
By substituting \(\lambda = 7\) and testing for intersection, we find they intersect.

Question. Find the equation of the line passing through the point \((-1, 3, -2)\) and perpendicular to the lines \(\frac{x}{1} = \frac{y}{2} = \frac{z}{3}\) and \(\frac{x+2}{-3} = \frac{y-1}{2} = \frac{z+1}{5}\).
Answer: Let the direction ratios of the required line be \(l, m, n\).
Since it is perpendicular to both lines:
\(l + 2m + 3n = 0\)
\(-3l + 2m + 5n = 0\)
Using cross-multiplication:
\(\frac{l}{10 - 6} = \frac{m}{-9 - 5} = \frac{n}{2 + 6} \Rightarrow \frac{l}{4} = \frac{m}{-14} = \frac{n}{8} \Rightarrow \frac{l}{2} = \frac{m}{-7} = \frac{n}{4}\).
Thus, the equation of the line through \((-1, 3, -2)\) is:
\(\frac{x+1}{2} = \frac{y-3}{-7} = \frac{z+2}{4}\).

Question. Find the distance between the point \((-1, -5, -10)\) and the point of intersection of the line \(\frac{x-2}{3} = \frac{y+1}{4} = \frac{z-2}{12}\) and the plane \(x - y + z = 5\).
Answer: Let any point on the line be \((3k+2, 4k-1, 12k+2)\).
Since this point lies on the plane \(x - y + z = 5\):
\((3k+2) - (4k-1) + (12k+2) = 5 \Rightarrow 11k + 5 = 5 \Rightarrow k = 0\).
So, the point of intersection is \((2, -1, 2)\).
The distance between \((-1, -5, -10)\) and \((2, -1, 2)\) is:
\(d = \sqrt{(2 - (-1))^2 + (-1 - (-5))^2 + (2 - (-10))^2} = \sqrt{9 + 16 + 144} = 13\) units.

Question. A plane makes intercepts \(-6, 3, 4\) respectively on the coordinate axes. Find the length of the perpendicular from the origin on it.
Answer: The equation of the plane is \(\frac{x}{-6} + \frac{y}{3} + \frac{z}{4} = 1 \Rightarrow -2x + 4y + 3z - 12 = 0\).
The length of the perpendicular from the origin \((0,0,0)\) is:
\(\frac{|-12|}{\sqrt{(-2)^2 + 4^2 + 3^2}} = \frac{12}{\sqrt{29}}\) unit.

Question. Prove that the line through \(A(0, -1, -1)\) and \(B(4, 5, 1)\) intersects the line through \(C(3, 9, 4)\) and \(D(-4, 4, 4)\).
Answer: Equation of line AB: \(\frac{x}{4} = \frac{y+1}{6} = \frac{z+1}{2} = \lambda \Rightarrow (4\lambda, 6\lambda - 1, 2\lambda - 1)\).
Equation of line CD: \(\frac{x-3}{-7} = \frac{y-9}{-5} = \frac{z-4}{0} = \mu \Rightarrow (-7\mu + 3, -5\mu + 9, 4)\).
Equating the coordinates:
\(2\lambda - 1 = 4 \Rightarrow \lambda = 5/2\).
\(4(5/2) = -7\mu + 3 \Rightarrow 10 = -7\mu + 3 \Rightarrow \mu = -1\).
Substituting \(\lambda = 5/2\) and \(\mu = -1\) into the y-coordinate equation:
\(6(5/2) - 1 = 14\) and \(-5(-1) + 9 = 14\).
Since the values of \(\lambda\) and \(\mu\) satisfy all three coordinate equations, the lines intersect.

Short Answer Type Questions - II

Question. Show that the lines \(\vec{r} = 3\hat{i} + 2\hat{j} - 4\hat{k} + \lambda(\hat{i} + 2\hat{j} + 2\hat{k})\) and \(\vec{r} = 5\hat{i} - 2\hat{j} + \mu(3\hat{i} + 2\hat{j} + 6\hat{k})\) are intersecting. Hence find their point of intersection.
Answer: Equating the coordinates of the general points of both lines:
\(3 + \lambda = 5 + 3\mu \Rightarrow \lambda - 3\mu = 2\)
\(2 + 2\lambda = -2 + 2\mu \Rightarrow \lambda - \mu = -2\)
\(-4 + 2\lambda = 6\mu \Rightarrow \lambda - 3\mu = 2\)
Solving the first two equations gives \(\lambda = -4\) and \(\mu = -2\). These values satisfy the third equation.
Therefore, the lines intersect. The point of intersection is \((-1, -6, -12)\).

Question. Using vectors, show that the points \(A(-2, 3, 5)\), \(B(7, 0, -1)\), \(C(-3, -2, -5)\) and \(D(3, 4, 7)\) are such that \(AB\) and \(CD\) intersect at the point \(P(1, 2, 3)\).
Answer: Let \(\vec{a} = -2\hat{i} + 3\hat{j} + 5\hat{k}\) and \(\vec{b} = 7\hat{i} - \hat{k}\).
The equation of the line joining \(A\) and \(B\) is \(\vec{r} = -2\hat{i} + 3\hat{j} + 5\hat{k} + \lambda(9\hat{i} - 3\hat{j} - 6\hat{k})\) ... (i)
Let \(\vec{c} = -3\hat{i} - 2\hat{j} - 5\hat{k}\) and \(\vec{d} = 3\hat{i} + 4\hat{j} + 7\hat{k}\).
The equation of the line joining \(C\) and \(D\) is \(\vec{r} = -3\hat{i} - 2\hat{j} - 5\hat{k} + \mu(6\hat{i} + 6\hat{j} + 12\hat{k})\) ... (ii)
Equating (i) and (ii) gives \(\lambda = 1/3\) and \(\mu = 2/3\).
Substituting \(\lambda = 1/3\) in (i), we obtain \(\vec{r} = \hat{i} + 2\hat{j} + 3\hat{k}\), which corresponds to the point \(P(1, 2, 3)\).

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: Let the direction ratios of the required line be \(l, m, n\). Since it is perpendicular to both given lines:
\(3l - 16m + 7n = 0\)
\(3l + 8m - 5n = 0\)
Using cross-multiplication:
\(\frac{l}{80 - 56} = \frac{m}{21 + 15} = \frac{n}{24 + 48} \Rightarrow \frac{l}{24} = \frac{m}{36} = \frac{n}{72} \Rightarrow \frac{l}{2} = \frac{m}{3} = \frac{n}{6}\).
Vector Equation: \(\vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + t(2\hat{i} + 3\hat{j} + 6\hat{k})\).
Cartesian Equation: \(\frac{x-1}{2} = \frac{y-2}{3} = \frac{z+4}{6}\).

Question. Find the shortest distance between the two lines whose vector equations are \(\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - 3\hat{j} + 2\hat{k})\) and \(\vec{r} = (4\hat{i} + 5\hat{j} + 6\hat{k}) + \mu(2\hat{i} + 3\hat{j} + \hat{k})\).
Answer: Here, \(\vec{a}_1 = \hat{i} + 2\hat{j} + 3\hat{k}\), \(\vec{b}_1 = \hat{i} - 3\hat{j} + 2\hat{k}\), and \(\vec{a}_2 = 4\hat{i} + 5\hat{j} + 6\hat{k}\), \(\vec{b}_2 = 2\hat{i} + 3\hat{j} + \hat{k}\).
\(\vec{a}_2 - \vec{a}_1 = 3\hat{i} + 3\hat{j} + 3\hat{k}\).
\(\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ 2 & 3 & 1 \end{vmatrix} = -9\hat{i} + 3\hat{j} + 9\hat{k}\).
\(|\vec{b}_1 \times \vec{b}_2| = \sqrt{(-9)^2 + 3^2 + 9^2} = 3\sqrt{19}\).
Now, \((\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = 3(-9) + 3(3) + 3(9) = 9\).
\(\therefore\) Shortest distance \(d = \frac{9}{3\sqrt{19}} = \frac{3}{\sqrt{19}}\) unit.

Question. Find the direction cosines of the line \(\frac{x+2}{2} = \frac{2y-7}{6} = \frac{5-z}{6}\). Also, find the vector equation of the line through the point \(A(-1, 2, 3)\) and parallel to the given line.
Answer: Rewriting the given line in standard form:
\(\frac{x+2}{2} = \frac{y - 7/2}{3} = \frac{z-5}{-6}\) ... (i)
Its direction ratios are \(2, 3, -6\).
As \(\sqrt{2^2 + 3^2 + (-6)^2} = 7\), its direction cosines are \(\frac{2}{7}, \frac{3}{7}, -\frac{6}{7}\).
The vector equation of the line through \(A(-1, 2, 3)\) parallel to (i) is:
\(\vec{r} = (-\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(2\hat{i} + 3\hat{j} - 6\hat{k})\).

Question. Find the shortest distance between the following 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: Comparing the equations, \(\vec{a}_1 = 2\hat{i} - 5\hat{j} + \hat{k}\), \(\vec{b}_1 = 3\hat{i} + 2\hat{j} + 6\hat{k}\) and \(\vec{a}_2 = 7\hat{i} - 6\hat{k}\), \(\vec{b}_2 = \hat{i} + 2\hat{j} + 2\hat{k}\).
\(\vec{a}_2 - \vec{a}_1 = 5\hat{i} + 5\hat{j} - 7\hat{k}\).
\(\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 6 \\ 1 & 2 & 2 \end{vmatrix} = -8\hat{i} + 4\hat{k}\).
\(|\vec{b}_1 \times \vec{b}_2| = \sqrt{(-8)^2 + 4^2} = 4\sqrt{5}\).
Shortest distance \(d = \frac{|(5\hat{i} + 5\hat{j} - 7\hat{k}) \cdot (-8\hat{i} + 4\hat{k})|}{4\sqrt{5}} = \frac{|-40 - 28|}{4\sqrt{5}} = \frac{68}{4\sqrt{5}} = \frac{17\sqrt{5}}{5}\) units.

Question. Find the shortest distance between the following lines: \(\frac{x+1}{7} = \frac{y+1}{-6} = \frac{z+1}{1}\) and \(\frac{x-3}{1} = \frac{y-5}{-2} = \frac{z-7}{1}\).
Answer: In vector form, the lines are:
\(\vec{r} = (-\hat{i} - \hat{j} - \hat{k}) + \lambda(7\hat{i} - 6\hat{j} + \hat{k})\)
\(\vec{r} = (3\hat{i} + 5\hat{j} + 7\hat{k}) + \mu(\hat{i} - 2\hat{j} + \hat{k})\)
\(\vec{a}_2 - \vec{a}_1 = 4\hat{i} + 6\hat{j} + 8\hat{k}\).
\(\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & -6 & 1 \\ 1 & -2 & 1 \end{vmatrix} = -4\hat{i} - 6\hat{j} - 8\hat{k}\).
Shortest distance \(d = \frac{|(-4\hat{i} - 6\hat{j} - 8\hat{k}) \cdot (4\hat{i} + 6\hat{j} + 8\hat{k})|}{\sqrt{(-4)^2 + (-6)^2 + (-8)^2}} = \frac{|-16 - 36 - 64|}{\sqrt{116}} = 2\sqrt{29}\) units.

Question. Find the unit vector perpendicular to the plane ABC where the position vectors of A, B and C are \(2\hat{i} - \hat{j} + \hat{k}\), \(\hat{i} + \hat{j} + 2\hat{k}\) and \(2\hat{i} + 3\hat{k}\) respectively.
Answer: \(\vec{AB} = (\hat{i} + \hat{j} + 2\hat{k}) - (2\hat{i} - \hat{j} + \hat{k}) = -\hat{i} + 2\hat{j} + \hat{k}\).
\(\vec{AC} = (2\hat{i} + 3\hat{k}) - (2\hat{i} - \hat{j} + \hat{k}) = \hat{j} + 2\hat{k}\).
A normal vector to the plane containing ABC is:
\(\vec{n} = \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 1 \\ 0 & 1 & 2 \end{vmatrix} = 3\hat{i} + 2\hat{j} - \hat{k}\).
The required unit vector is \(\frac{\vec{n}}{|\vec{n}|} = \frac{3\hat{i} + 2\hat{j} - \hat{k}}{\sqrt{9 + 4 + 1}} = \frac{1}{\sqrt{14}}(3\hat{i} + 2\hat{j} - \hat{k})\).

Question. Find the vector equation of the plane through the points \((2, 1, -1)\) and \((-1, 3, 4)\) and perpendicular to the plane \(x - 2y + 4z = 10\).
Answer: Let the equation of the plane through \((2, 1, -1)\) be \(a(x - 2) + b(y - 1) + c(z + 1) = 0\) ... (i)
Since the point \((-1, 3, 4)\) lies on it:
\(-3a + 2b + 5c = 0\) ... (ii)
Also, (i) is perpendicular to \(x - 2y + 4z = 10\):
\(a(1) - 2b + 4c = 0 \Rightarrow a - 2b + 4c = 0\) ... (iii)
Solving (ii) and (iii) gives \(\frac{a}{18} = \frac{b}{17} = \frac{c}{4}\).
Thus, the equation of the plane is \(18(x-2) + 17(y-1) + 4(z+1) = 0 \Rightarrow 18x + 17y + 4z = 49\).
In vector form: \(\vec{r} \cdot (18\hat{i} + 17\hat{j} + 4\hat{k}) = 49\).

Question. Show that the lines \(\frac{5-x}{-4} = \frac{y-7}{4} = \frac{z+3}{-5}\) and \(\frac{x-8}{7} = \frac{2y-8}{2} = \frac{z-5}{3}\) are coplanar.
Answer: The lines can be written as:
\(\frac{x-5}{4} = \frac{y-7}{4} = \frac{z+3}{-5}\) and \(\frac{x-8}{7} = \frac{y-4}{1} = \frac{z-5}{3}\).
The coplanarity condition is \(\begin{vmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0\).
L.H.S = \begin{vmatrix} 8-5 & 4-7 & 5 - (-3) \\ 4 & 4 & -5 \\ 7 & 1 & 3 \end{vmatrix} = \begin{vmatrix} 3 & -3 & 8 \\ 4 & 4 & -5 \\ 7 & 1 & 3 \end{vmatrix} = 3(12 + 5) + 3(12 + 35) + 8(4 - 28) = 51 + 141 - 192 = 0 = R.H.S.
Hence, the given lines are coplanar.

Question. Find the image of the point having position vector \(\hat{i} + 3\hat{j} + 4\hat{k}\) in the plane \(\vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) + 3 = 0\).
Answer: Point \(P \equiv (1, 3, 4)\) and the plane equation is \(2x - y + z + 3 = 0\) ... (i)
Any line perpendicular to (i) passing through \(P\) is \(\frac{x-1}{2} = \frac{y-3}{-1} = \frac{z-4}{1} = \lambda\).
Let \(M(2\lambda + 1, -\lambda + 3, \lambda + 4)\) be the point of intersection with the plane.
Substituting \(M\) in (i) gives \(2(2\lambda + 1) - (-\lambda + 3) + (\lambda + 4) + 3 = 0 \Rightarrow 6\lambda + 6 = 0 \Rightarrow \lambda = -1\).
So, \(M \equiv (-1, 4, 3)\).
Let \(Q(\alpha, \beta, \gamma)\) be the image of \(P\). Then \(M\) is the midpoint of \(PQ\):
\(\frac{\alpha + 1}{2} = -1 \Rightarrow \alpha = -3\); \(\frac{\beta + 3}{2} = 4 \Rightarrow \beta = 5\); \(\frac{\gamma + 4}{2} = 3 \Rightarrow \gamma = 2\).
Hence, the image of \(P\) is \(Q(-3, 5, 2)\).

Question. Find the equation of the plane passing through the points \((-1, 2, 0)\), \((2, 2, -1)\) and parallel to the line \(\frac{x-1}{1} = \frac{2y+1}{2} = \frac{z+1}{-1}\).
Answer: Let the plane equation be \(a(x + 1) + b(y - 2) + cz = 0\) ... (i)
Since it passes through \((2, 2, -1)\): \(3a - c = 0\) ... (ii)
Since it is parallel to the line \(\frac{x-1}{1} = \frac{y + 1/2}{1} = \frac{z+1}{-1}\):
\(a(1) + b(1) + c(-1) = 0 \Rightarrow a + b - c = 0\) ... (iii)
From (ii) and (iii), we get \(a = \lambda\), \(b = 2\lambda\), and \(c = 3\lambda\).
Substituting these in (i) gives: \(x + 2y + 3z = 3\).

Question. Find the equation of a plane which passes through the point \((3, 2, 0)\) and contains the line \(\frac{x-3}{1} = \frac{y-6}{5} = \frac{z-4}{4}\).
Answer: Let the plane equation be \(a(x - 3) + b(y - 2) + cz = 0\) ... (i)
Since the plane contains the line \(\frac{x-3}{1} = \frac{y-6}{5} = \frac{z-4}{4}\):
\(a(3 - 3) + b(6 - 2) + c(4 - 0) = 0 \Rightarrow 4b + 4c = 0\)
and \(a(1) + b(5) + c(4) = 0 \Rightarrow a + 5b + 4c = 0\).
Solving these gives \(a = -4\lambda\), \(b = 4\lambda\), \(c = -4\lambda\).
Substituting into (i): \(-4\lambda(x-3) + 4\lambda(y-2) - 4\lambda z = 0 \Rightarrow x - y + z - 1 = 0\).

Question. Find the coordinates of the point where the line through the points \(A(3, 4, 1)\) and \(B(5, 1, 6)\) crosses the XY-plane.
Answer: The line through \(A\) and \(B\) is \(\frac{x-3}{2} = \frac{y-4}{-3} = \frac{z-1}{5}\) ... (i)
This line meets the XY-plane (where \(z = 0\)):
\(\frac{x-3}{2} = \frac{y-4}{-3} = \frac{-1}{5} \Rightarrow x = 3 - \frac{2}{5} = \frac{13}{5}\) and \(y = 4 + \frac{3}{5} = \frac{23}{5}\).
Thus, the point is \(\left(\frac{13}{5}, \frac{23}{5}, 0\right)\).

Long Answer Type Questions

Question. Find the coordinates of the foot of perpendicular and the length of the perpendicular drawn from the point \(P(5, 4, 2)\) to the line, \(\vec{r} = -\hat{i} + 3\hat{j} + \hat{k} + \lambda(2\hat{i} + 3\hat{j} - \hat{k})\). Also find the image of \(P\) in this line.
Answer: The Cartesian equation of the line is \(\frac{x+1}{2} = \frac{y-3}{3} = \frac{z-1}{-1} = \lambda\).
Let the foot of the perpendicular be \(Q(2\lambda - 1, 3\lambda + 3, -\lambda + 1)\).
The direction ratios of \(PQ\) are \((2\lambda - 6, 3\lambda - 1, -\lambda - 1)\).
Since \(PQ\) is perpendicular to the line:
\(2(2\lambda - 6) + 3(3\lambda - 1) - 1(-\lambda - 1) = 0 \Rightarrow 14\lambda - 14 = 0 \Rightarrow \lambda = 1\).
Thus, \(Q\) is \((1, 6, 0)\), which is the foot of the perpendicular.
Length of perpendicular \(PQ = \sqrt{(1-5)^2 + (6-4)^2 + (0-2)^2} = \sqrt{16 + 4 + 4} = 2\sqrt{6}\) units.
If \(R(\alpha, \beta, \gamma)\) is the image of \(P\):
\(\frac{\alpha+5}{2} = 1 \Rightarrow \alpha = -3\); \(\frac{\beta+4}{2} = 6 \Rightarrow \beta = 8\); \(\frac{\gamma+2}{2} = 0 \Rightarrow \gamma = -2\).
Thus, the image is \(R(-3, 8, -2)\).

Question. Find the distance of the point \((2, 12, 5)\) from the point of intersection of the line \(\vec{r} = 2\hat{i} - 4\hat{j} + 2\hat{k} + \lambda(3\hat{i} + 4\hat{j} + 2\hat{k})\) and the plane \(\vec{r} \cdot (\hat{i} - 2\hat{j} + \hat{k}) = 0\).
Answer: Let a general point on the line be \((3\lambda + 2, 4\lambda - 4, 2\lambda + 2)\).
Since this point lies on the plane \(x - 2y + z = 0\):
\((3\lambda + 2) - 2(4\lambda - 4) + (2\lambda + 2) = 0 \Rightarrow -3\lambda + 12 = 0 \Rightarrow \lambda = 4\).
The point of intersection is \((14, 12, 10)\).
Its distance from \((2, 12, 5)\) is:
\(d = \sqrt{(14-2)^2 + (12-12)^2 + (10-5)^2} = \sqrt{144 + 25} = 13\) units.

Question. Find the vector equation of a line passing through the point \((2, 3, 2)\) and parallel to the line \(\vec{r} = (-2\hat{i} + 3\hat{j}) + \lambda(2\hat{i} - 3\hat{j} + 6\hat{k})\). Also, find the distance between these two lines.
Answer: The vector equation of the parallel line is \(\vec{r} = (2\hat{i} + 3\hat{j} + 2\hat{k}) + \mu(2\hat{i} - 3\hat{j} + 6\hat{k})\).
Here, \(\vec{a}_1 = -2\hat{i} + 3\hat{j}\), \(\vec{a}_2 = 2\hat{i} + 3\hat{j} + 2\hat{k}\), and \(\vec{b} = 2\hat{i} - 3\hat{j} + 6\hat{k}\).
\(\vec{a}_2 - \vec{a}_1 = 4\hat{i} + 2\hat{k}\).
\(\vec{b} \times (\vec{a}_2 - \vec{a}_1) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 6 \\ 4 & 0 & 2 \end{vmatrix} = -6\hat{i} + 20\hat{j} + 12\hat{k}\).
Distance \(d = \frac{|\vec{b} \times (\vec{a}_2 - \vec{a}_1)|}{|\vec{b}|} = \frac{\sqrt{(-6)^2 + 20^2 + 12^2}}{\sqrt{4 + 9 + 36}} = \frac{\sqrt{580}}{7}\) units.

Question. Show that the lines \(\frac{x-2}{1} = \frac{y-2}{3} = \frac{z-3}{1}\) and \(\frac{x-2}{1} = \frac{y-3}{4} = \frac{z-4}{2}\) intersect. Also, find the coordinates of the point of intersection. Find the equation of the plane containing the two lines.
Answer: Any point on the first line is \((r+2, 3r+2, r+3)\) and on the second line is \((k+2, 4k+3, 2k+4)\).
For intersection, equating coordinates gives \(r = k = -1\).
The point of intersection is \((1, -1, 2)\).
The equation of the plane containing both lines is:
\(\begin{vmatrix} x-2 & y-2 & z-3 \\ 1 & 3 & 1 \\ 1 & 4 & 2 \end{vmatrix} = 0 \Rightarrow 2x - y + z - 5 = 0\).

Question. Find the vector and cartesian equations of a plane containing the two lines \(\vec{r} = 2\hat{i} + 3\hat{j} - 3\hat{k} + \lambda(\hat{i} + 2\hat{j} + 5\hat{k})\) and \(\vec{r} = 3\hat{i} + 3\hat{j} + 2\hat{k} + \mu(3\hat{i} - 2\hat{j} + 5\hat{k})\). Also show that the line \(\vec{r} = (2\hat{i} + 5\hat{j} + 2\hat{k}) + p(3\hat{i} - 2\hat{j} + 5\hat{k})\) lies in the plane.
Answer: The plane containing the lines passes through \(\vec{a}_1 = 2\hat{i} + 3\hat{j} - 3\hat{k}\) and is parallel to \(\vec{b}_1 = \hat{i} + 2\hat{j} + 5\hat{k}\) and \(\vec{b}_2 = 3\hat{i} - 2\hat{j} + 5\hat{k}\).
The normal to the plane is:
\(\vec{n} = \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 5 \\ 3 & -2 & 5 \end{vmatrix} = 20\hat{i} + 10\hat{j} - 8\hat{k} \Rightarrow 10\hat{i} + 5\hat{j} - 4\hat{k}\).
The vector equation is: \(\vec{r} \cdot (10\hat{i} + 5\hat{j} - 4\hat{k}) = 37\).
The Cartesian equation is: \(10x + 5y - 4z = 37\).
For the given line \(\vec{r} = (2\hat{i} + 5\hat{j} + 2\hat{k}) + p(3\hat{i} - 2\hat{j} + 5\hat{k})\) to lie in the plane:
The point \((2, 5, 2)\) must satisfy the plane: \(10(2) + 5(5) - 4(2) = 20 + 25 - 8 = 37\).
The direction of the line must be perpendicular to the plane's normal: \(3(10) - 2(5) + 5(-4) = 30 - 10 - 20 = 0\).
Since both conditions are satisfied, the line lies completely in the plane.