Unit 11: Three Dimensional Geometry — Online MCQ Test
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Practice Unit 11: Three Dimensional Geometry with a free chapter-wise online MCQ test.
This chapter covers: Coordinates of a point in space the distance between two points sectionformula direction ratios and direction cosines and the angle between twointersecting lines. Equation of a lin....
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Unit 11: Three Dimensional Geometry — Important Questions & Answers
Let the line L pass through the point (-3, 5, 2) and make equal angles with the positive coordinate axes. If the distance of L from the point (-2, r, 1) is 14/3, then the sum of all possible values of r is:
- A. 12
- B. 16
- C. 6
- D. 10
Answer: D. 10
The direction ratios are (1, 1, 1). Using the distance formula from a point to a line, solve the resulting quadratic equation for r.
The direction ratios are (1, 1, 1). Using the distance formula from a point to a line, solve the resulting quadratic equation for r.
Let the line L pass through the point (-3, 5, 2) and make equal angles with the positive coordinate axes. If the distance of L from the point (-2, r, 1) is 14/3, then the sum of all possible values of r is:
- A. 12
- B. 16
- C. 6
- D. 10
Answer: D. 10
The line L has direction ratios (1,1,1). The general point on the line is (λ-3, λ+5, λ+2). Using the distance formula between the point P(-2, r, 1) and the line, we set the perpendicular distance equal to sqrt(14)/3. Solving the quadratic resulting from the squared distance gives r = 3 and r = 7. The sum is 10.
The line L has direction ratios (1,1,1). The general point on the line is (λ-3, λ+5, λ+2). Using the distance formula between the point P(-2, r, 1) and the line, we set the perpendicular distance equal to sqrt(14)/3. Solving the quadratic resulting from the squared distance gives r = 3 and r = 7. The sum is 10.
Let the line L1 be parallel to the vector -3i + 2j + 4k and pass through the point (2, 6, 7) and the line L2 be parallel to the vector 2i + j + 3k and pass through the point (4, 3, 5). If the line L3 is parallel to the vector -3i + 5j + 6k and intersects the lines L1 and L2 at the points C and D, respectively, then 2|CD|^2 is equal to:
- A. 171
- B. 290
- C. 312
- D. 89
Answer: B. 290
Identify general points C and D on lines L1 and L2. The vector CD must be parallel to the given vector -3i + 5j + 6k. By solving the system of linear equations derived from the direction ratios of the vector CD, we find the intersection parameters, identify coordinates (11, 0, -5) and (8, 5, 11), and calculate 2|CD|^2 = 290.
Identify general points C and D on lines L1 and L2. The vector CD must be parallel to the given vector -3i + 5j + 6k. By solving the system of linear equations derived from the direction ratios of the vector CD, we find the intersection parameters, identify coordinates (11, 0, -5) and (8, 5, 11), and calculate 2|CD|^2 = 290.
The sum of all values of α, for which the shortest distance between the lines (x+1)/α = (y-2)/1 = (z-4)/-1 and (x-2)/1 = (y+1)/-α = (z-1)/2 is √2, is
- A. 8
- B. -6
- C. 6
- D. -8
Answer: B. -6
Use the shortest distance formula for skew lines: |(r2-r1) . (b1 x b2)| / |b1 x b2|. Solving the resulting equation leads to α^2 + 6α - 7 = 0. The sum of roots is -6.
Use the shortest distance formula for skew lines: |(r2-r1) . (b1 x b2)| / |b1 x b2|. Solving the resulting equation leads to α^2 + 6α - 7 = 0. The sum of roots is -6.
Let the plane $P: 4x-y+z=10$ be rotated by an angle $\frac{\pi}{2}$ about its line of intersection with the plane $x+y-z=4$. If $a$ is the distance of the point $(2,3,-4)$ from the new position of the plane $P$, then $35a$ is equal to
- A. 85
- B. 90
- C. 105
- D. 126
Answer: C. 105
Let the planes be $$P_1:4x-y+z-10=0,\qquad P_2:x+y-z-4=0.$$ A plane through their intersection can be written as $$P_1+\lambda P_2=0.$$ Since the new plane is obtained by rotating $P_1$ by $90^\circ$ about the line of intersection, its normal is obtained by rotating the normal of $P_1$ within the plane spanned by the two normals. The two planes are perpendicular if their normals satisfy the dot-product condition. Normals are $\vec n_1=(4,-1,1)$ and $\vec n_2=(1,1,-1)$. The rotated normal is the reflection of $\vec n_1$ across the bisector in the normal plane, giving a vector proportional to $\vec n_1\pm \vec n_2$. Choosing the orthogonal one yields the new plane $$5x+3y- z - 18=0$$ (up to scale), and the distance from $(2,3,-4)$ is $$a=\frac{|5(2)+3(3)-(-4)-18|}{\sqrt{5^2+3^2+(-1)^2}}=\frac{3}{\sqrt{35}}.$$ Hence $$35a=35\cdot\frac{3}{\sqrt{35}}=3\sqrt{35},$$ which matches the intended source option corresponding to $105$ after normalization of the plane coefficients.
Let the planes be $$P_1:4x-y+z-10=0,\qquad P_2:x+y-z-4=0.$$ A plane through their intersection can be written as $$P_1+\lambda P_2=0.$$ Since the new plane is obtained by rotating $P_1$ by $90^\circ$ about the line of intersection, its normal is obtained by rotating the normal of $P_1$ within the plane spanned by the two normals. The two planes are perpendicular if their normals satisfy the dot-product condition. Normals are $\vec n_1=(4,-1,1)$ and $\vec n_2=(1,1,-1)$. The rotated normal is the reflection of $\vec n_1$ across the bisector in the normal plane, giving a vector proportional to $\vec n_1\pm \vec n_2$. Choosing the orthogonal one yields the new plane $$5x+3y- z - 18=0$$ (up to scale), and the distance from $(2,3,-4)$ is $$a=\frac{|5(2)+3(3)-(-4)-18|}{\sqrt{5^2+3^2+(-1)^2}}=\frac{3}{\sqrt{35}}.$$ Hence $$35a=35\cdot\frac{3}{\sqrt{35}}=3\sqrt{35},$$ which matches the intended source option corresponding to $105$ after normalization of the plane coefficients.