Question 6 of 13intermediate🔧 ApplyLong Answer4 marks

Show that the points (−2,3,5)(-2, 3, 5), (1,2,3)(1, 2, 3) and (7,0,−1)(7, 0, -1) are collinear.

Correct Answer

Let the given points be P(-2, 3, 5), Q(1, 2, 3), and R(7, 0, -1).

Points are said to be collinear if they lie on a line. Three points are collinear if the sum of distances between two consecutive points equals the distance between the extreme points.

Distance PQ:

PQ=(1+2)2+(2−3)2+(3−5)2PQ = \sqrt{(1 + 2)^2 + (2 - 3)^2 + (3 - 5)^2} =(3)2+(−1)2+(−2)2= \sqrt{(3)^2 + (-1)^2 + (-2)^2} =9+1+4=14= \sqrt{9 + 1 + 4} = \sqrt{14}

Distance QR:

QR=(7−1)2+(0−2)2+(−1−3)2QR = \sqrt{(7 - 1)^2 + (0 - 2)^2 + (-1 - 3)^2} =(6)2+(−2)2+(−4)2= \sqrt{(6)^2 + (-2)^2 + (-4)^2} =36+4+16=56=214= \sqrt{36 + 4 + 16} = \sqrt{56} = 2\sqrt{14}

Distance PR:

PR=(7+2)2+(0−3)2+(−1−5)2PR = \sqrt{(7 + 2)^2 + (0 - 3)^2 + (-1 - 5)^2} =(9)2+(−3)2+(−6)2= \sqrt{(9)^2 + (-3)^2 + (-6)^2} =81+9+36=126=314= \sqrt{81 + 9 + 36} = \sqrt{126} = 3\sqrt{14}

Verification:

PQ+QR=14+214=314=PRPQ + QR = \sqrt{14} + 2\sqrt{14} = 3\sqrt{14} = PR

Since PQ + QR = PR, the points P, Q, and R are collinear.

Exercise: EXERCISE 11.2 | Q: 2 | (Page Page 6)
For More Understanding

Explanation

This question tests the understanding of collinearity in three-dimensional geometry. The textbook demonstrates that three points are collinear when the sum of distances between consecutive points equals the total distance between the extreme points. The solution uses the 3D distance formula and shows that PQ (14\sqrt{14}) + QR (2142\sqrt{14}) = PR (3143\sqrt{14}), proving collinearity.

Solution Steps

  1. Step 1: Identify points P(-2, 3, 5), Q(1, 2, 3), R(7, 0, -1)

  2. Step 2: Calculate PQ using distance formula = 14\sqrt{14}

  3. Step 3: Calculate QR using distance formula = 2142\sqrt{14}

  4. Step 4: Calculate PR using distance formula = 3143\sqrt{14}

  5. Step 5: Verify PQ + QR = PR to prove collinearity

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