Mastering 3D Geometry: Lines and Planes
This quiz focuses on the core concepts of Vector and 3D Geometry, specifically prioritizing the shortest distance between lines and the intersection of planes. These topics are fundamental for advanced spatial reasoning in mathematics.
- Understanding skew and parallel lines in three-dimensional space.
- Calculating distances from points to planes and between parallel planes.
- Solving systems of equations to find points and lines of intersection.
- Applying vector dot and cross products to geometric problems.
Designed for a moderate difficulty level, these 25 questions will test your ability to apply vector formulas and coordinate geometry to solve complex spatial problems.
Q1. Find the shortest distance between the skew 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}##.
The shortest distance ##d## between skew lines is given by ##\frac{\mid (\vec{a_2}-\vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) \mid}{\mid \vec{b_1} \times \vec{b_2} \mid}##. Here ##\vec{b_1} \times \vec{b_2} = (-1, 2, -1)## and ##\vec{a_2}-\vec{a_1} = (1, 2, 2)##. Dot product is ##-1+4-2=1##. Magnitude is ##\sqrt{6}##. Distance = ##1/\sqrt{6}##.
Q2. What is the direction ratio of the line of intersection of the planes ##x + y + z = 1## and ##2x + 3y - z = 4##?
The direction vector of the line of intersection is the cross product of the normal vectors of the two planes. ##\vec{n_1} = (1, 1, 1)## and ##\vec{n_2} = (2, 3, -1)##. ##\vec{n_1} \times \vec{n_2} = (-1-3, 2+1, 3-2) = (-4, 3, 1)##.
Q3. Find the point of intersection of the three planes: ##x+y+z=6##, ##x-y+z=2##, and ##2x+y-z=1##.
By solving the system: (1)+(2) gives ##2x+2z=8 \implies x+z=4##. (2)+(3) gives ##3x=3 \implies x=1##. Substituting ##x=1## into ##x+z=4## gives ##z=3##. Substituting ##x=1, z=3## into ##x+y+z=6## gives ##y=2##. The point is ##(1, 2, 3)##.
Q4. Find the shortest 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})##.
For parallel lines, distance ##d = \frac{\mid \vec{b} \times (\vec{a_2}-\vec{a_1}) \mid}{\mid \vec{b} \mid}##. Here ##\vec{a_2}-\vec{a_1} = (2, 1, 1)## and ##\vec{b} = (2, 3, 6)##. ##\vec{b} \times (\vec{a_2}-\vec{a_1}) = (-3, 10, -4)##. Magnitude is ##\sqrt{9+100+16} = \sqrt{125} = 5\sqrt{5}##. ##\mid \vec{b} \mid = 7##. Distance = ##5\sqrt{5}/7##.
Q5. The distance of the point ##(1, 1, 1)## from the plane ##x - y + z + 5 = 0## is:
Distance ##d = \frac{\mid ax_1+by_1+cz_1+d \mid}{\sqrt{a^2+b^2+c^2}} = \frac{\mid 1-1+1+5 \mid}{\sqrt{1+1+1}} = \frac{6}{\sqrt{3}} = 2\sqrt{3}##.
Q6. Find the equation of the plane passing through the intersection of ##x + y + z = 1## and ##2x + 3y - z + 4 = 0## and the point ##(1, 1, 1)##.
The family of planes is ##(x+y+z-1) + k(2x+3y-z+4) = 0##. Substituting ##(1,1,1)##: ##(1+1+1-1) + k(2+3-1+4) = 2 + 8k = 0 \implies k = -1/4##. The equation is ##4(x+y+z-1) - (2x+3y-z+4) = 0 \implies 2x + y + 5z - 8 = 0##.
Q7. The angle between the planes ##2x - y + z = 6## and ##x + y + 2z = 3## is:
The angle ##\theta## is given by ##\cos \theta = \frac{\mid \vec{n_1} \cdot \vec{n_2} \mid}{\mid \vec{n_1} \mid \mid \vec{n_2} \mid}##. ##\vec{n_1} = (2, -1, 1)##, ##\vec{n_2} = (1, 1, 2)##. ##\vec{n_1} \cdot \vec{n_2} = 2 - 1 + 2 = 3##. ##\mid \vec{n_1} \mid = \sqrt{6}##, ##\mid \vec{n_2} \mid = \sqrt{6}##. ##\cos \theta = 3/6 = 1/2 \implies \theta = 60^\circ##.
Q8. If the lines ##\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-a}{4}## and ##\frac{x-4}{5} = \frac{y-1}{2} = \frac{z}{1}## intersect, find the value of ##a##.
For intersection, the shortest distance must be zero. The condition is ##\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \end{vmatrix} = 0##. ##\begin{vmatrix} 3 & -1 & -a \\ 2 & 3 & 4 \\ 5 & 2 & 1 \end{vmatrix} = 3(3-8) + 1(2-20) - a(4-15) = -15 - 18 + 11a = 0 \implies 11a = 33 \implies a = 3##.
Q9. Find the foot of the perpendicular from the origin to the plane ##2x + 3y + 4z - 29 = 0##.
The foot of the perpendicular ##(x, y, z)## from ##(x_1, y_1, z_1)## is ##\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} = -\frac{ax_1+by_1+cz_1+d}{a^2+b^2+c^2}##. For origin: ##\frac{x}{2} = \frac{y}{3} = \frac{z}{4} = -\frac{-29}{4+9+16} = \frac{29}{29} = 1##. So ##x=2, y=3, z=4##.
Q10. Find the shortest distance between the lines ##\vec{r} = \lambda(\hat{i}+2\hat{j}+\hat{k})## and ##\vec{r} = (\hat{i}+\hat{j}+\hat{k}) + \mu(2\hat{i}+\hat{j}-\hat{k})##.
##\vec{a_2}-\vec{a_1} = (1,1,1)##. ##\vec{b_1} \times \vec{b_2} = (-3, 3, -3)##. Dot product = ##-3+3-3 = -3##. Magnitude of cross product = ##\sqrt{9+9+9} = 3\sqrt{3}##. Distance = ##\mid -3 \mid / 3\sqrt{3} = 1/\sqrt{3}##.
Q11. Find the point of intersection of the line ##\vec{r} = (2\hat{i}-\hat{j}+2\hat{k}) + \lambda(3\hat{i}+4\hat{j}+2\hat{k})## and the plane ##\vec{r} \cdot (\hat{i}-\hat{j}+\hat{k}) = 5##.
Substitute the line equation into the plane equation: ##((2+3\lambda)\hat{i} + (-1+4\lambda)\hat{j} + (2+2\lambda)\hat{k}) \cdot (\hat{i}-\hat{j}+\hat{k}) = 5##. ##(2+3\lambda) - (-1+4\lambda) + (2+2\lambda) = 5 \implies 5 + \lambda = 5 \implies \lambda = 0##. The point is ##(2, -1, 2)##.
Q12. Find the equation of the plane containing the line ##\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}## and parallel to the x-axis.
The plane is parallel to the x-axis ##(1,0,0)## and contains the line direction ##(2,3,4)##. The normal to the plane is ##(2,3,4) \times (1,0,0) = (0, 4, -3)##. The plane passes through ##(1,2,3)##. Equation: ##0(x-1) + 4(y-2) - 3(z-3) = 0 \implies 4y - 8 - 3z + 9 = 0 \implies 4y - 3z + 1 = 0##.
Q13. Find the shortest distance between lines ##\frac{x-3}{3} = \frac{y-8}{-1} = \frac{z-3}{1}## and ##\frac{x+3}{-3} = \frac{y+7}{2} = \frac{z-6}{4}##.
##\vec{a_2}-\vec{a_1} = (-6, -15, 3)##. ##\vec{b_1} \times \vec{b_2} = (-6, -15, 3)##. Dot product = ##(-6)^2 + (-15)^2 + 3^2 = 270##. Magnitude ##\mid \vec{b_1} \times \vec{b_2} \mid = \sqrt{270}##. Distance = ##270 / \sqrt{270} = \sqrt{270} = 3\sqrt{30}##.
Q14. Find the distance between the parallel planes ##2x - y + 2z + 3 = 0## and ##4x - 2y + 4z + 15 = 0##.
Divide the second plane by 2: ##2x - y + 2z + 7.5 = 0##. Distance ##d = \frac{\mid d_2 - d_1 \mid}{\sqrt{a^2+b^2+c^2}} = \frac{\mid 7.5 - 3 \mid}{\sqrt{4+1+4}} = \frac{4.5}{3} = 1.5 = 3/2##.
Q15. A plane contains two intersecting lines with direction ratios ##(2, 3, 4)## and ##(-1, 2, 1)##. What are the direction ratios of the normal to this plane?
The normal vector is the cross product of the two direction vectors. ##(2, 3, 4) \times (-1, 2, 1) = (3-8, -4-2, 4+3) = (-5, -6, 7)##.
Q16. Find the image of the point ##(1, 3, 4)## in the plane ##2x - y + z + 3 = 0##.
Using the image formula: ##\frac{x-1}{2} = \frac{y-3}{-1} = \frac{z-4}{1} = -2\frac{2(1)-3+4+3}{4+1+1} = -2(6/6) = -2##. Solving gives ##x = 1-4 = -3##, ##y = 3+2 = 5##, ##z = 4-2 = 2##. Image is ##(-3, 5, 2)##.
Q17. The lines ##\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}## and ##\frac{x-4}{5} = \frac{y-1}{2} = \frac{z-k}{1}## are coplanar. Find ##k##.
Condition for coplanarity: ##\begin{vmatrix} 3 & -1 & k-3 \\ 2 & 3 & 4 \\ 5 & 2 & 1 \end{vmatrix} = 0##. Expanding gives ##3(3-8) + 1(2-20) + (k-3)(4-15) = -15 - 18 - 11(k-3) = -33 - 11k + 33 = 0 \implies k=0##.
Q18. Find the distance of the origin from the plane passing through ##(1, 0, 0)##, ##(0, 2, 0)##, and ##(0, 0, 3)##.
The intercept form of the plane is ##x/1 + y/2 + z/3 = 1 \implies 6x + 3y + 2z - 6 = 0##. Distance from origin = ##\mid -6 \mid / \sqrt{36+9+4} = 6/\sqrt{49} = 6/7##.
Q19. Find the angle between the line ##\frac{x-1}{1} = \frac{y-2}{2} = \frac{z-3}{2}## and the plane ##2x + y - 2z + 5 = 0##.
##\sin \theta = \frac{\mid \vec{b} \cdot \vec{n} \mid}{\mid \vec{b} \mid \mid \vec{n} \mid}##. ##\vec{b} = (1, 2, 2)##, ##\vec{n} = (2, 1, -2)##. ##\vec{b} \cdot \vec{n} = 2 + 2 - 4 = 0##. Since ##\sin \theta = 0##, ##\theta = 0^\circ##. The line is parallel to the plane.
Q20. Find the vector equation of the line of intersection of the planes ##\vec{r} \cdot (\hat{i}+\hat{j}+\hat{k}) = 1## and ##\vec{r} \cdot (\hat{i}+2\hat{j}+3\hat{k}) = 4##.
Direction of line = ##(1,1,1) \times (1,2,3) = (1,-2,1)##. To find a point, set ##z=0## in both plane equations: ##x+y=1## and ##x+2y=4##. Solving gives ##y=3, x=-2##. Point is ##(-2,3,0)##. Equation is ##\vec{r} = (-2\hat{i}+3\hat{j}) + \lambda(\hat{i}-2\hat{j}+\hat{k})##.
Q21. What is the shortest distance between the z-axis and the line ##x = 1, z = 0##?
The z-axis is the line ##x=0, y=0##. The given line is parallel to the y-axis (since ##x## and ##z## are constant). The shortest distance between ##x=0, z=0## and ##x=1, z=0## is simply the difference in the x-coordinates, which is 1.
Q22. Find the equation of the plane passing through the origin and parallel to the lines with direction ratios ##(1, 2, 3)## and ##(2, 3, 1)##.
The normal to the plane is the cross product of the two direction vectors: ##(1, 2, 3) \times (2, 3, 1) = (2-9, 6-1, 3-4) = (-7, 5, -1)##. Since it passes through the origin, the equation is ##-7x + 5y - z = 0##, which is ##7x - 5y + z = 0##.
Q23. If a line is parallel to a plane, what is the relationship between the direction vector of the line ##\vec{b}## and the normal vector of the plane ##\vec{n}##?
If a line is parallel to a plane, the line is perpendicular to the normal of the plane. Therefore, their dot product must be zero.
Q24. Find the symmetric form of the line of intersection of the planes ##x+y-z=1## and ##2x-3y+z=2##.
The direction vector is ##(1,1,-1) \times (2,-3,1) = (-2,-3,-5)## or ##(2,3,5)##. A point on the line (setting ##z=0##) is ##(1,0,0)##. Both forms represent the same line with different sign conventions for direction ratios.
Q25. The shortest distance between two lines is given as zero. This implies that the lines are:
A shortest distance of zero means the lines either intersect at a point or are the same line (coincident), both of which imply they lie in the same plane (coplanar).
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