Introduction to Three Dimensional Geometry — Practice

Question 1

A point has coordinates (0, -5, 3). Which coordinate plane does it lie in, and which octant would a point with all three coordinates negated from this one lie in?

Accepted Answer

Since the x-coordinate is 0, the point lies in the YZ-plane. Negating all three coordinates gives (0,5,-3), which still has x=0, so it lies on the boundary between octants, not strictly inside one. Answer: The original point lies in the YZ-plane.

Question 2

Name the octant in which the point (-2, 5, -8) lies.

Accepted Answer

Signs are (-,+,-). Answer: Comparing with the sign table, this is octant VI.

Question 3

Find the distance between the points A(1, 2, 3) and B(4, 6, 3).

Accepted Answer

AB = sqrt[(4-1)^2+(6-2)^2+(3-3)^2] = sqrt[9+16+0] = sqrt(25) = 5. Answer: The distance is 5 units.

Question 4

Show that the points (2, -1, 3), (5, -5, 12), and (-1, 3, -6) are collinear.

Accepted Answer

Labelling the points P(2,-1,3), Q(5,-5,12), R(-1,3,-6): PQ=sqrt[(5-2)^2+(-5+1)^2+(12-3)^2]=sqrt[9+16+81]=sqrt(106). PR=sqrt[(-1-2)^2+(3+1)^2+(-6-3)^2]=sqrt[9+16+81]=sqrt(106). QR=sqrt[(-1-5)^2+(3+5)^2+(-6-12)^2]=sqrt[36+64+324]=sqrt(424)=2sqrt(106). Since PQ+PR=sqrt(106)+sqrt(106)=2sqrt(106)=QR, point P lies between Q and R. Answer: Since PQ+PR=QR, the three points are collinear.

Question 5

Two vertices of a triangle are (3, -2, 4) and (-1, 6, 0), and its centroid is (2, 1, 3). Find the third vertex.

Accepted Answer

Let the third vertex be (x,y,z). Using the centroid formula: (3-1+x)/3=2 gives x=4; (-2+6+y)/3=1 gives y=-1; (4+0+z)/3=3 gives z=5. Answer: The third vertex is (4, -1, 5).

Question 6

Find the midpoint of the segment joining (2, -3, 6) and (-4, 5, 0).

Accepted Answer

Averaging coordinates: ((2-4)/2, (-3+5)/2, (6+0)/2) = (-1, 1, 3). Answer: The midpoint is (-1, 1, 3).

Question 7

Find the distance of the point (3, -4, 5) from the origin.

Accepted Answer

Using OQ = sqrt(x^2+y^2+z^2): sqrt(9+16+25) = sqrt(50) = 5sqrt(2). Answer: The distance from the origin is 5sqrt(2) units.

Question 8

Show that the points A(1, 2, 3), B(-1, -2, -1), and C(2, 3, 2) form a right-angled triangle by computing all three side lengths.

Accepted Answer

AB^2 = (-1-1)^2+(-2-2)^2+(-1-3)^2 = 4+16+16 = 36. BC^2 = (2+1)^2+(3+2)^2+(2+1)^2 = 9+25+9 = 43. AC^2 = (2-1)^2+(3-2)^2+(2-3)^2 = 1+1+1 = 3. If the triangle were right-angled, the right angle would sit opposite the longest side, BC, so the Pythagorean relation to check is AB^2+AC^2=BC^2. Here AB^2+AC^2=36+3=39, which is not equal to BC^2=43. Answer: Since AB^2+AC^2 does not equal BC^2, the triangle is scalene but not right-angled.

Practice — Introduction to Three Dimensional Geometry

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