Solved Examples

Important Theorems and Formulas Recap

1. Direction Cosines and Ratios

• For direction cosines $l, m, n$, the fundamental identity is $l^2 + m^2 + n^2 = 1$.
• If $a, b, c$ are direction ratios, then the corresponding direction cosines are $$l = \pm \frac{a}{\sqrt{a^2+b^2+c^2}}, \quad m = \pm \frac{b}{\sqrt{a^2+b^2+c^2}}, \quad n = \pm \frac{c}{\sqrt{a^2+b^2+c^2}}$$ where the same sign is taken throughout, depending on the orientation of the directed line.
• Direction ratios of a line joining $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ are $$(x_2-x_1), \ (y_2-y_1), \ (z_2-z_1).$$

2. Equation of a Line

• Passing through position vector $\vec{a}$ and parallel to vector $\vec{b}$: $$\vec{r} = \vec{a} + \lambda \vec{b}$$ and in Cartesian form, $$\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}.$$
• Passing through points with position vectors $\vec{a}$ and $\vec{b}$: $$\vec{r} = \vec{a} + \lambda(\vec{b}-\vec{a}).$$

3. Shortest Distance Between Lines

• For skew lines: $$d = \left| \frac{(\vec{a}_2-\vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|} \right|$$
• For parallel lines: $$d = \frac{|(\vec{a}_2-\vec{a}_1) \times \vec{b}|}{|\vec{b}|}.$$

4. Equation of a Plane

• Normal form: $$\vec{r} \cdot \hat{n} = d$$ or equivalently $$lx + my + nz = d,$$ where $d$ is the perpendicular distance from the origin.
• Passing through point $\vec{a}$ with normal $\vec{N}$: $$(\vec{r}-\vec{a}) \cdot \vec{N} = 0.$$
• Coplanarity condition for two lines: $$(\vec{a}_2-\vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = 0.$$

5. Distances and Angles

• Angle between lines: $$\cos\theta = \left| \frac{\vec{b}_1 \cdot \vec{b}_2}{|\vec{b}_1||\vec{b}_2|} \right|$$
• Angle between planes: $$\cos\theta = \left| \frac{\vec{n}_1 \cdot \vec{n}_2}{|\vec{n}_1||\vec{n}_2|} \right|$$
• Angle between a line and a plane: $$\sin\phi = \left| \frac{\vec{b} \cdot \vec{n}}{|\vec{b}||\vec{n}|} \right|$$
• Distance from point $(x_1,y_1,z_1)$ to plane $Ax+By+Cz+D=0$: $$d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2+B^2+C^2}}.$$

Example 1: Direction Cosines

If a line makes equal angles with the x, y, and z-axes, find its direction cosines.

Solution: Step 1: If the line makes equal angles with all three axes, let each angle be $\alpha$. Then $$\alpha = \beta = \gamma.$$ Hence the three direction cosines are equal: $$l = m = n = \cos\alpha.$$

Step 2: Use the fundamental identity for direction cosines: $$l^2 + m^2 + n^2 = 1.$$ Since $l=m=n$, we get $$3l^2 = 1.$$ Thus, $$l^2 = \frac{1}{3}.$$

Step 3: Therefore, $$l = \pm \frac{1}{\sqrt{3}}.$$ Since all three are equal, the direction cosines are either all positive or all negative.

Answer: $$\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right) \text{ or } \left(-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\right).$$

Example 2: Direction Ratios from Points

Find the direction cosines of the line passing through the points $(-2, 4, -5)$ and $(1, 2, 3)$.

Solution: Step 1: Let the points be $$P(-2,4,-5), \quad Q(1,2,3).$$ The direction ratios of the line joining $P$ to $Q$ are obtained by coordinate differences: $$a = 1-(-2)=3, \quad b = 2-4=-2, \quad c = 3-(-5)=8.$$ So the direction ratios are $$(3,-2,8).$$

Step 2: Find the magnitude of this direction vector: $$\sqrt{a^2+b^2+c^2} = \sqrt{3^2+(-2)^2+8^2} = \sqrt{9+4+64} = \sqrt{77}.$$

Step 3: Divide each direction ratio by the magnitude to get direction cosines: $$l = \frac{3}{\sqrt{77}}, \quad m = -\frac{2}{\sqrt{77}}, \quad n = \frac{8}{\sqrt{77}}.$$

Answer: $$\frac{3}{\sqrt{77}},\ -\frac{2}{\sqrt{77}},\ \frac{8}{\sqrt{77}}$$

Example 3: Equation of a Line

Find the Cartesian equation of the line which passes through the point $(-2, 4, -5)$ and is parallel to the vector $3\hat{i} + 5\hat{j} + 6\hat{k}$.

Solution: Step 1: The line passes through the fixed point $$(x_1,y_1,z_1)=(-2,4,-5).$$

Step 2: Since the line is parallel to $$3\hat{i}+5\hat{j}+6\hat{k},$$ its direction ratios are $$a=3, \quad b=5, \quad c=6.$$

Step 3: Use the Cartesian form of the line: $$\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}.$$ Substituting the point and direction ratios, $$\frac{x-(-2)}{3} = \frac{y-4}{5} = \frac{z-(-5)}{6}.$$ So, $$\frac{x+2}{3} = \frac{y-4}{5} = \frac{z+5}{6}.$$

Answer: $$\frac{x+2}{3} = \frac{y-4}{5} = \frac{z+5}{6}$$

Example 4: Line through Two Points

Find the vector equation for the line passing through the points $(3, -2, -5)$ and $(3, -2, 6)$.

Solution: Step 1: Write the position vectors of the two points: $$\vec{a} = 3\hat{i} - 2\hat{j} - 5\hat{k}, \quad \vec{b} = 3\hat{i} - 2\hat{j} + 6\hat{k}.$$

Step 2: Find the direction vector: $$\vec{b} - \vec{a} = (3-3)\hat{i} + (-2+2)\hat{j} + (6-(-5))\hat{k} = 11\hat{k}.$$ So the line is parallel to the z-axis.

Step 3: Use the two-point vector form: $$\vec{r} = \vec{a} + \lambda(\vec{b}-\vec{a}).$$ Therefore, $$\vec{r} = (3\hat{i} - 2\hat{j} - 5\hat{k}) + \lambda(11\hat{k}).$$ Since $11\lambda$ is again an arbitrary scalar, we may write this more simply as $$\vec{r} = 3\hat{i} - 2\hat{j} - 5\hat{k} + \mu\hat{k}.$$

Answer: $$\vec{r} = 3\hat{i} - 2\hat{j} - 5\hat{k} + \mu\hat{k}$$

Example 5: Angle Between Two Lines

Find the angle between the lines $\vec{r} = 3\hat{i} + 2\hat{j} - 4\hat{k} + \lambda(\hat{i} + 2\hat{j} + 2\hat{k})$ and $\vec{r} = 5\hat{i} - 2\hat{j} + \mu(3\hat{i} + 2\hat{j} + 6\hat{k})$.

Solution: Step 1: Extract the direction vectors of the two lines: $$\vec{b}_1 = \hat{i} + 2\hat{j} + 2\hat{k}, \quad \vec{b}_2 = 3\hat{i} + 2\hat{j} + 6\hat{k}.$$

Step 2: Find the dot product: $$\vec{b}_1 \cdot \vec{b}_2 = (1)(3) + (2)(2) + (2)(6) = 3 + 4 + 12 = 19.$$

Step 3: Find the magnitudes: $$|\vec{b}_1| = \sqrt{1^2+2^2+2^2} = \sqrt{9} = 3,$$ $$|\vec{b}_2| = \sqrt{3^2+2^2+6^2} = \sqrt{49} = 7.$$

Step 4: Use the angle formula between two lines: $$\cos\theta = \left|\frac{\vec{b}_1\cdot\vec{b}_2}{|\vec{b}_1||\vec{b}_2|}\right| = \frac{19}{3\cdot 7} = \frac{19}{21}.$$ Therefore, $$\theta = \cos^{-1}\left(\frac{19}{21}\right).$$

Answer: $$\cos^{-1}(19/21)$$

Example 6: Perpendicular Lines Condition

Find $p$ so that lines $\frac{x-1}{3} = \frac{y-2}{2p/7} = \frac{z-3}{2}$ and $\frac{x-1}{-3p/7} = \frac{y-5}{1} = \frac{z-6}{-5}$ are perpendicular.

Solution: Step 1: Extract the direction ratios of the two lines: For the first line, $$a_1 = 3, \quad b_1 = \frac{2p}{7}, \quad c_1 = 2.$$ For the second line, $$a_2 = -\frac{3p}{7}, \quad b_2 = 1, \quad c_2 = -5.$$

Step 2: Two lines are perpendicular if the dot product of their direction vectors is zero: $$a_1a_2 + b_1b_2 + c_1c_2 = 0.$$ So, $$3\left(-\frac{3p}{7}\right) + \left(\frac{2p}{7}\right)(1) + 2(-5) = 0.$$

Step 3: Simplify: $$-\frac{9p}{7} + \frac{2p}{7} - 10 = 0$$ $$-\frac{7p}{7} - 10 = 0$$ $$-p - 10 = 0.$$ Hence, $$p = -10.$$

Answer: $$-10$$

Example 7: Shortest Distance Between Skew Lines

Find the shortest distance between the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and $\frac{x-2}{3} = \frac{y-4}{4} = \frac{z-5}{5}$.

Solution: Step 1: Identify points and direction vectors: For the first line, $$\vec{a}_1 = (1,2,3), \quad \vec{b}_1 = (2,3,4).$$ For the second line, $$\vec{a}_2 = (2,4,5), \quad \vec{b}_2 = (3,4,5).$$

Step 2: Compute the vector joining the chosen points: $$\vec{a}_2 - \vec{a}_1 = (1,2,2).$$

Step 3: Compute the cross product of the direction vectors: $$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix}.$$ Expanding, $$\vec{b}_1 \times \vec{b}_2 = \hat{i}(15-16) - \hat{j}(10-12) + \hat{k}(8-9) = -\hat{i} + 2\hat{j} - \hat{k}.$$

Step 4: Find its magnitude: $$|\vec{b}_1 \times \vec{b}_2| = \sqrt{(-1)^2 + 2^2 + (-1)^2} = \sqrt{6}.$$

Step 5: Find the scalar triple product numerator: $$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (1,2,2) \cdot (-1,2,-1).$$ Thus, $$-1 + 4 - 2 = 1.$$

Step 6: Therefore the shortest distance is $$d = \left|\frac{1}{\sqrt{6}}\right| = \frac{1}{\sqrt{6}}.$$

Answer: $$\frac{1}{\sqrt{6}}$$

Example 8: Distance Between Parallel Lines

Find the distance between the parallel lines $\vec{r} = \hat{i} + 2\hat{j} - 4\hat{k} + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})$ and $\vec{r} = 3\hat{i} + 3\hat{j} - 5\hat{k} + \mu(2\hat{i} + 3\hat{j} + 6\hat{k})$.

Solution: Step 1: Identify $$\vec{a}_1 = (1,2,-4), \quad \vec{a}_2 = (3,3,-5), \quad \vec{b} = (2,3,6).$$ Since both lines have the same direction vector, they are parallel.

Step 2: Find the vector joining the chosen points: $$\vec{a}_2 - \vec{a}_1 = (2,1,-1).$$

Step 3: Compute the cross product: $$(\vec{a}_2-\vec{a}_1) \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 2 & 3 & 6 \end{vmatrix}.$$ Expanding, $$= \hat{i}(6-(-3)) - \hat{j}(12-(-2)) + \hat{k}(6-2)$$ $$= 9\hat{i} - 14\hat{j} + 4\hat{k}.$$

Step 4: Find its magnitude: $$|(\vec{a}_2-\vec{a}_1) \times \vec{b}| = \sqrt{9^2 + (-14)^2 + 4^2} = \sqrt{81+196+16} = \sqrt{293}.$$

Step 5: Find the magnitude of the common direction vector: $$|\vec{b}| = \sqrt{2^2+3^2+6^2} = \sqrt{49} = 7.$$

Step 6: Therefore, $$d = \frac{\sqrt{293}}{7}.$$

Answer: $$\frac{\sqrt{293}}{7}$$

Example 9: Equation of a Plane (Normal Form)

Find the vector equation of a plane which is at a distance of $6/\sqrt{29}$ from the origin and its normal vector from the origin is $2\hat{i} - 3\hat{j} + 4\hat{k}$.

Solution: Step 1: The given normal vector is $$\vec{N} = 2\hat{i} - 3\hat{j} + 4\hat{k}.$$ Its magnitude is $$|\vec{N}| = \sqrt{2^2+(-3)^2+4^2} = \sqrt{4+9+16} = \sqrt{29}.$$

Step 2: Therefore the unit normal vector is $$\hat{n} = \frac{\vec{N}}{|\vec{N}|} = \frac{2\hat{i} - 3\hat{j} + 4\hat{k}}{\sqrt{29}}.$$

Step 3: Since the plane is at distance $$d = \frac{6}{\sqrt{29}}$$ from the origin, its normal form is $$\vec{r} \cdot \hat{n} = d.$$ So, $$\vec{r} \cdot \left(\frac{2\hat{i} - 3\hat{j} + 4\hat{k}}{\sqrt{29}}\right) = \frac{6}{\sqrt{29}}.$$

Step 4: Multiply both sides by $\sqrt{29}$: $$\vec{r} \cdot (2\hat{i} - 3\hat{j} + 4\hat{k}) = 6.$$

Answer: $$\vec{r} \cdot (2\hat{i} - 3\hat{j} + 4\hat{k}) = 6$$

Example 10: Plane Passing Through a Point

Find the Cartesian equation of the plane passing through $(1, 2, -4)$ and parallel to the plane $x - 2y + 4z = 3$.

Solution: Step 1: A plane parallel to $$x - 2y + 4z = 3$$ has the same normal vector. So its equation must be of the form $$x - 2y + 4z = d.$$

Step 2: Substitute the given point $(1,2,-4)$ into this equation: $$1 - 2(2) + 4(-4) = d.$$ So, $$1 - 4 - 16 = d \implies d = -19.$$

Step 3: Therefore the required plane is $$x - 2y + 4z = -19.$$ Writing in standard form, $$x - 2y + 4z + 19 = 0.$$

Answer: $$x - 2y + 4z + 19 = 0$$

Example 11: Intercept Form

Find the intercepts cut off by the plane $2x + y - z = 5$.

Solution: Step 1: Divide the plane equation by 5 to make the right-hand side equal to 1: $$\frac{2x}{5} + \frac{y}{5} - \frac{z}{5} = 1.$$

Step 2: Rewrite in intercept form: $$\frac{x}{5/2} + \frac{y}{5} + \frac{z}{-5} = 1.$$ Comparing with $$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1,$$ we get $$a = \frac{5}{2}, \quad b = 5, \quad c = -5.$$

Answer: $$\frac{5}{2},\ 5,\ -5$$

Example 12: Equation of Plane Through Three Points

Find the equation of the plane through the points $(1,1,0)$, $(1,2,1),$ and $(-2,2,-1)$.

Solution: Step 1: Let the three points be $$A(1,1,0), \quad B(1,2,1), \quad C(-2,2,-1).$$ Use the determinant form of the plane through three points: $$\begin{vmatrix} x-1 & y-1 & z-0 \\ 1-1 & 2-1 & 1-0 \\ -2-1 & 2-1 & -1-0 \end{vmatrix} = 0.$$ This becomes $$\begin{vmatrix} x-1 & y-1 & z \\ 0 & 1 & 1 \\ -3 & 1 & -1 \end{vmatrix} = 0.$$

Step 2: Expand along the first row: $$(x-1)\begin{vmatrix}1 & 1 \\ 1 & -1\end{vmatrix} - (y-1)\begin{vmatrix}0 & 1 \\ -3 & -1\end{vmatrix} + z\begin{vmatrix}0 & 1 \\ -3 & 1\end{vmatrix} = 0.$$

Step 3: Evaluate the minors: $$(x-1)(-1-1) - (y-1)(0-(-3)) + z(0-(-3)) = 0.$$ So, $$-2(x-1) - 3(y-1) + 3z = 0.$$ Expand: $$-2x + 2 - 3y + 3 + 3z = 0$$ $$-2x - 3y + 3z + 5 = 0.$$ Multiplying by $-1$, $$2x + 3y - 3z = 5.$$

Answer: $$2x + 3y - 3z = 5$$

Example 13: Plane Through Intersection

Find the equation of the plane passing through the intersection of $x+y+z=6$ and $2x+3y+4z+5=0$ and the point $(1,1,1)$.

Solution: Step 1: Write the family of planes through the line of intersection: $$(x+y+z-6) + \lambda(2x+3y+4z+5) = 0.$$

Step 2: Since the required plane passes through $(1,1,1)$, substitute these coordinates: $$(1+1+1-6) + \lambda(2+3+4+5) = 0.$$ This gives $$-3 + 14\lambda = 0,$$ so $$\lambda = \frac{3}{14}.$$

Step 3: Substitute this value of $\lambda$ into the family: $$(x+y+z-6) + \frac{3}{14}(2x+3y+4z+5)=0.$$ Multiply throughout by 14: $$14(x+y+z-6) + 3(2x+3y+4z+5)=0.$$ Expand: $$14x+14y+14z-84 + 6x+9y+12z+15 = 0.$$ So, $$20x + 23y + 26z - 69 = 0.$$

Answer: $$20x + 23y + 26z = 69$$

Example 14: Coplanarity of Lines

Check if lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and $\frac{x-4}{5} = \frac{y-1}{2} = \frac{z}{1}$ are coplanar.

Solution: Step 1: From the first line, $$(x_1,y_1,z_1) = (1,2,3), \quad (a_1,b_1,c_1) = (2,3,4).$$ From the second line, $$(x_2,y_2,z_2) = (4,1,0), \quad (a_2,b_2,c_2) = (5,2,1).$$

Step 2: Use the coplanarity determinant: $$\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0.$$ Substitute values: $$\begin{vmatrix} 3 & -1 & -3 \\ 2 & 3 & 4 \\ 5 & 2 & 1 \end{vmatrix}.$$

Step 3: Expand: $$3(3\cdot1 - 4\cdot2) - (-1)(2\cdot1 - 4\cdot5) + (-3)(2\cdot2 - 3\cdot5).$$ That is, $$3(3-8) + (2-20) - 3(4-15)$$ $$= 3(-5) - 18 - 3(-11)$$ $$= -15 - 18 + 33 = 0.$$

Since the determinant is zero, the lines are coplanar.

Answer: Yes, the lines are coplanar.

Example 15: Angle Between Planes

Find the angle between the planes $x+y+z=1$ and $x-2y+3z=1$.

Solution: Step 1: The normal vectors of the two planes are $$\vec{n}_1 = (1,1,1), \quad \vec{n}_2 = (1,-2,3).$$

Step 2: Compute the dot product: $$\vec{n}_1 \cdot \vec{n}_2 = 1(1) + 1(-2) + 1(3) = 2.$$

Step 3: Compute the magnitudes: $$|\vec{n}_1| = \sqrt{1+1+1} = \sqrt{3},$$ $$|\vec{n}_2| = \sqrt{1+4+9} = \sqrt{14}.$$

Step 4: Use the angle formula between planes: $$\cos\theta = \left|\frac{\vec{n}_1 \cdot \vec{n}_2}{|\vec{n}_1||\vec{n}_2|}\right| = \frac{2}{\sqrt{3}\sqrt{14}} = \frac{2}{\sqrt{42}}.$$ Hence, $$\theta = \cos^{-1}\left(\frac{2}{\sqrt{42}}\right).$$

Answer: $$\cos^{-1}(2/\sqrt{42})$$

Example 16: Angle Between Line and Plane

Find the angle between the line $\frac{x+1}{2} = \frac{y}{3} = \frac{z-3}{6}$ and the plane $10x + 2y - 11z = 3$.

Solution: Step 1: The direction vector of the line is $$\vec{b} = (2,3,6),$$ and the normal vector of the plane is $$\vec{n} = (10,2,-11).$$

Step 2: Compute the dot product: $$\vec{b}\cdot\vec{n} = 2(10) + 3(2) + 6(-11) = 20 + 6 - 66 = -40.$$

Step 3: Compute the magnitudes: $$|\vec{b}| = \sqrt{4+9+36} = 7,$$ $$|\vec{n}| = \sqrt{100+4+121} = \sqrt{225} = 15.$$

Step 4: Use the formula for the angle $\phi$ between a line and a plane: $$\sin\phi = \left|\frac{\vec{b}\cdot\vec{n}}{|\vec{b}||\vec{n}|}\right| = \frac{40}{7\cdot15} = \frac{40}{105} = \frac{8}{21}.$$ Thus, $$\phi = \sin^{-1}\left(\frac{8}{21}\right).$$

Answer: $$\sin^{-1}(8/21)$$

Example 17: Distance of Point from Plane

Find the distance of the point $(2, 5, -3)$ from the plane $\vec{r} \cdot (6\hat{i} - 3\hat{j} + 2\hat{k}) = 4$.

Solution: Step 1: Convert the vector equation into Cartesian form: $$6x - 3y + 2z = 4 \implies 6x - 3y + 2z - 4 = 0.$$ So, $$A=6, \quad B=-3, \quad C=2, \quad D=-4.$$ The point is $$(2,5,-3).$$

Step 2: Use the distance formula: $$d = \frac{|6(2) - 3(5) + 2(-3) - 4|}{\sqrt{6^2 + (-3)^2 + 2^2}}.$$

Step 3: Simplify the numerator: $$|12 - 15 - 6 - 4| = |-13| = 13.$$

Step 4: Simplify the denominator: $$\sqrt{36+9+4} = \sqrt{49} = 7.$$

Step 5: Therefore, $$d = \frac{13}{7}.$$

Answer: $$\frac{13}{7}$$

Example 18: Distance Between Parallel Planes

Find the distance between the parallel planes $2x - y + 3z + 4 = 0$ and $2x - y + 3z - 10 = 0$.

Solution: Step 1: Since the coefficients of $x,y,z$ are the same, the planes are parallel. Use the formula $$d = \frac{|d_1-d_2|}{\sqrt{A^2+B^2+C^2}}.$$ Here, $$A=2, \quad B=-1, \quad C=3, \quad d_1=4, \quad d_2=-10.$$

Step 2: Compute the numerator: $$|4 - (-10)| = 14.$$

Step 3: Compute the denominator: $$\sqrt{2^2 + (-1)^2 + 3^2} = \sqrt{4+1+9} = \sqrt{14}.$$

Step 4: Therefore, $$d = \frac{14}{\sqrt{14}} = \sqrt{14}.$$

Answer: $$\sqrt{14}$$

Example 19: Foot of the Perpendicular

Find the coordinates of the foot of the perpendicular from $(1, 1, 1)$ to the plane $x + 2y + 3z = 2$.

Solution: Step 1: The normal vector of the plane is $$(1,2,3).$$ So the perpendicular line through $(1,1,1)$ is $$\frac{x-1}{1} = \frac{y-1}{2} = \frac{z-1}{3} = \lambda.$$

Step 2: A general point on this line is $$Q(\lambda+1,\ 2\lambda+1,\ 3\lambda+1).$$

Step 3: Since $Q$ lies on the plane, substitute into $$x+2y+3z=2.$$ So, $$(\lambda+1) + 2(2\lambda+1) + 3(3\lambda+1) = 2.$$ Simplify: $$\lambda+1 + 4\lambda+2 + 9\lambda+3 = 2$$ $$14\lambda + 6 = 2$$ $$14\lambda = -4$$ $$\lambda = -\frac{2}{7}.$$

Step 4: Substitute back: $$x = 1-\frac{2}{7} = \frac{5}{7},$$ $$y = 1-\frac{4}{7} = \frac{3}{7},$$ $$z = 1-\frac{6}{7} = \frac{1}{7}.$$

Answer: $$\left(\frac{5}{7},\frac{3}{7},\frac{1}{7}\right)$$

JEE Main Standard

Example 20: Image of a Point in a Plane

Find the image of the point $(3, 1, 2)$ in the plane $2x - y + z + 4 = 0$.

Solution: Step 1: The normal vector of the plane is $$(2,-1,1).$$ Hence the line through $(3,1,2)$ perpendicular to the plane is $$\frac{x-3}{2} = \frac{y-1}{-1} = \frac{z-2}{1} = \lambda.$$ A general point on this line is $$Q(2\lambda+3,\ -\lambda+1,\ \lambda+2).$$

Step 2: Since $Q$ is the foot of the perpendicular, it must lie on the plane: $$2(2\lambda+3) - (-\lambda+1) + (\lambda+2) + 4 = 0.$$ Simplify: $$4\lambda+6 + \lambda-1 + \lambda+2 + 4 = 0$$ $$6\lambda + 11 = 0$$ $$\lambda = -\frac{11}{6}.$$

Step 3: Therefore the foot of the perpendicular is $$Q\left(2\left(-\frac{11}{6}\right)+3,\ -\left(-\frac{11}{6}\right)+1,\ -\frac{11}{6}+2\right)$$ $$= \left(-\frac{2}{3},\ \frac{17}{6},\ \frac{1}{6}\right).$$

Step 4: Let the image point be $P'(x',y',z')$. Since $Q$ is the midpoint of $PP'$, we have $$x' = 2Q_x - 3, \quad y' = 2Q_y - 1, \quad z' = 2Q_z - 2.$$ So, $$x' = 2\left(-\frac{2}{3}\right) - 3 = -\frac{4}{3} - 3 = -\frac{13}{3},$$ $$y' = 2\left(\frac{17}{6}\right) - 1 = \frac{17}{3} - 1 = \frac{14}{3},$$ $$z' = 2\left(\frac{1}{6}\right) - 2 = \frac{1}{3} - 2 = -\frac{5}{3}.$$

Answer: $$\left(-\frac{13}{3},\ \frac{14}{3},\ -\frac{5}{3}\right)$$

Example 21: Line Intersection with Plane

Find the distance of the point $(-1, -5, -10)$ from the point of intersection of the line $\frac{x-2}{3} = \frac{y+1}{4} = \frac{z-2}{12}$ and the plane $x - y + z = 5$.

Solution: Step 1: Write the general point on the line by taking the common parameter as $\lambda$: $$x = 3\lambda + 2, \quad y = 4\lambda - 1, \quad z = 12\lambda + 2.$$ So the general point is $$Q(3\lambda+2,\ 4\lambda-1,\ 12\lambda+2).$$

Step 2: Since the point of intersection lies on the plane $x-y+z=5$, substitute: $$(3\lambda+2) - (4\lambda-1) + (12\lambda+2) = 5.$$ Simplify: $$3\lambda + 2 - 4\lambda + 1 + 12\lambda + 2 = 5$$ $$11\lambda + 5 = 5$$ $$\lambda = 0.$$

Step 3: Therefore the intersection point is $$Q(2,-1,2).$$

Step 4: Find the distance from $P(-1,-5,-10)$ to $Q(2,-1,2)$: $$PQ = \sqrt{(2+1)^2 + (-1+5)^2 + (2+10)^2}$$ $$= \sqrt{3^2 + 4^2 + 12^2} = \sqrt{9+16+144} = \sqrt{169} = 13.$$

Answer: $$13$$

Example 22: Shortest Distance with Parameter

If the shortest distance between the lines $\frac{x-1}{\alpha} = \frac{y+1}{-1} = \frac{z}{1}$ and $\frac{x+1}{2} = \frac{y-2}{1} = \frac{z-1}{2}$ is $1/\sqrt{3}$, find $\alpha$.

Solution: Step 1: Write the two lines in vector-data form. For the first line, $$\vec{a}_1 = (1,-1,0), \quad \vec{b}_1 = (\alpha,-1,1).$$ For the second line, $$\vec{a}_2 = (-1,2,1), \quad \vec{b}_2 = (2,1,2).$$

Step 2: Compute the vector joining the chosen points: $$\vec{a}_2 - \vec{a}_1 = (-2,3,1).$$

Step 3: Compute the cross product: $$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \alpha & -1 & 1 \\ 2 & 1 & 2 \end{vmatrix}.$$ Expanding, $$\vec{b}_1 \times \vec{b}_2 = \hat{i}((-1)(2)-1(1)) - \hat{j}(\alpha\cdot2 - 1\cdot2) + \hat{k}(\alpha\cdot1 - (-1)\cdot2).$$ Thus, $$\vec{b}_1 \times \vec{b}_2 = (-3,\ 2-2\alpha,\ \alpha+2).$$

Step 4: Compute the numerator of the shortest distance formula: $$(\vec{a}_2-\vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)$$ $$= (-2,3,1)\cdot(-3,\ 2-2\alpha,\ \alpha+2)$$ $$= 6 + 3(2-2\alpha) + (\alpha+2)$$ $$= 6 + 6 - 6\alpha + \alpha + 2 = 14 - 5\alpha.$$ So the numerator is $$|14 - 5\alpha|.$$

Step 5: Compute the denominator: $$|\vec{b}_1 \times \vec{b}_2| = \sqrt{(-3)^2 + (2-2\alpha)^2 + (\alpha+2)^2}.$$ This becomes $$\sqrt{9 + (4 - 8\alpha + 4\alpha^2) + (\alpha^2 + 4\alpha + 4)}$$ $$= \sqrt{5\alpha^2 - 4\alpha + 17}.$$

Step 6: Use the given shortest distance: $$\frac{|14-5\alpha|}{\sqrt{5\alpha^2 - 4\alpha + 17}} = \frac{1}{\sqrt{3}}.$$ Squaring both sides, $$\frac{(14-5\alpha)^2}{5\alpha^2 - 4\alpha + 17} = \frac{1}{3}.$$ So, $$3(14-5\alpha)^2 = 5\alpha^2 - 4\alpha + 17.$$ Expanding, $$3(196 - 140\alpha + 25\alpha^2) = 5\alpha^2 - 4\alpha + 17$$ $$588 - 420\alpha + 75\alpha^2 = 5\alpha^2 - 4\alpha + 17.$$ Rearranging, $$70\alpha^2 - 416\alpha + 571 = 0.$$

Step 7: Solve the quadratic: $$\alpha = \frac{416 \pm \sqrt{416^2 - 4\cdot70\cdot571}}{140}.$$ This gives $$\alpha = \frac{104}{35} \pm \frac{3\sqrt{366}}{70}.$$

Answer: $$\alpha = \frac{104}{35} \pm \frac{3\sqrt{366}}{70}$$

Example 23: Plane Containing Two Lines

Find the equation of the plane containing the lines $\vec{r} = \hat{i} + \hat{j} + \lambda(\hat{i} + 2\hat{j} - \hat{k})$ and $\vec{r} = \hat{i} + \hat{j} + \mu(-\hat{i} + \hat{j} - 2\hat{k})$.

Solution: Step 1: Both lines pass through the common point $$(1,1,0).$$ So this point lies on the required plane.

Step 2: The direction vectors of the two lines are $$\vec{b}_1 = (1,2,-1), \quad \vec{b}_2 = (-1,1,-2).$$

Step 3: A normal vector to the plane is $$\vec{n} = \vec{b}_1 \times \vec{b}_2.$$ Compute: $$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -1 \\ -1 & 1 & -2 \end{vmatrix}$$ $$= \hat{i}((-4)-(-1)) - \hat{j}((-2)-1) + \hat{k}(1-(-2))$$ $$= -3\hat{i} + 3\hat{j} + 3\hat{k}.$$ This simplifies to the normal vector $$(1,-1,-1).$$

Step 4: Use the point-normal form through $(1,1,0)$: $$1(x-1) - 1(y-1) - 1(z-0) = 0.$$ Simplifying, $$x - 1 - y + 1 - z = 0$$ $$x - y - z = 0.$$

Answer: $$x - y - z = 0$$