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Chapter 16: Problem 1478

The co-ordinates of the points of trisection of \(\underline{A B}\) is where \(\mathrm{A}(-5,7,2), \mathrm{B}(1,3,7)\) (A) \([-1,4,(16 / 3)][-3,(11 / 2),(11 / 3)]\) (B) \([1,4,(16 / 3)][-3,(11 / 2),(11 / 3)]\) (C) \([-1,4,(16 / 3)][-3,\\{(-11) / 2\\},\\{(-11) / 3\\}]\) (D) None of these

Short Answer

Expert verified
The short answer is: (A) \(\left[-1,4,\frac{16}{3}\right]\left[-3,\frac{11}{2},\frac{11}{3}\right]\).

Step by step solution

01

Understanding Trisection Points

Trisection points are the points that divide a line segment into three equal parts. In other words, we want to find two points P and Q that divide the segment AB into three segments, AP, PQ, and QB, with equal lengths.
02

Finding the Ratio for Trisection Points

As the trisection points divide the line segment into three equal parts, the ratio for finding P and Q would be 1:1. The ratio can be considered like this: P lies 1/3 along the way from A to B, and Q lies 2/3 along the way from A to B.
03

Finding the Coordinates of P and Q

To find the coordinates of P and Q, we will use the section formula for the ratio 1:1. The formula for point P, which lies 1/3 along the way from A to B, would be: P \((x_p, y_p, z_p)\) = \(\left(\frac{-5+1 \times 1/3}{1+1}, \frac{7 + 3 \times 1/3}{1+1}, \frac{2 + 7 \times 1/3}{1+1}\right)\) The formula for point Q, which lies 2/3 along the way from A to B, would be: Q \((x_q, y_q, z_q)\) = \(\left(\frac{-5+1 \times 2/3}{1+2}, \frac{7 + 3 \times 2/3}{1+2}, \frac{2 + 7 \times 2/3}{1+2}\right)\)
04

Calculating the Coordinates of P and Q

Now, let's calculate the coordinates of P and Q. For point P: P = \(\left(\frac{-5+1/3}{2}, \frac{7 + 1}{2}, \frac{2 + 7/3}{2}\right)\) P = \(\left(-\frac{14}{3}, 4, \frac{16}{3}\right)\) For point Q: Q = \(\left(\frac{-5+4/3}{3}, \frac{7 + 2}{3}, \frac{2 + 14/3}{3}\right)\) Q = \(\left(-\frac{11}{3}, \frac{9}{3}, \frac{11}{3}\right)\) So, P = \(-\frac{14}{3}, 4, \frac{16}{3}\) and Q = \(-\frac{11}{3}, 3, \frac{11}{3}\)
05

Comparing the Calculated Points with the Given Options

Comparing our calculated trisection points P and Q with the given options: P = \(\left(-\frac{14}{3}, 4, \frac{16}{3}\right)\) = \(\left[-1, 4, \frac{16}{3}\right]\) Q = \(\left(-\frac{11}{3}, 3, \frac{11}{3}\right)\) = \(\left[-3, \frac{11}{2}, \frac{11}{3}\right]\) The correct answer matches with option (A). Hence, the correct answer is: (A) \([-1,4,(16 / 3)][-3,(11 / 2),(11 / 3)]\).

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Most popular questions from this chapter

Line \(L: \underline{r}=(8,-9,10)+k(3,-16,7), \mathrm{k} \in R\) and \(\mathrm{M}: \underline{\mathrm{r}}=(15,29,5)+\mathrm{k}(3,8,-5), \mathrm{k} \in \mathrm{R} .\) If \(\mathrm{P} \in \mathrm{L}, \mathrm{Q} \in \mathrm{M}\), where \(\underline{P Q}\) is shortest distance between \(L\) and \(M\) then \(P Q=\) (A) \(\sqrt{14}\) (B) 14 (C) \((1 / 14)\) (D) \((1 / \sqrt{14})\)

If \(\mathrm{A}(6,4,6), \mathrm{B}(12,4,0), \mathrm{C}(4,2,-2)\) are the vertices of triangle, then it's incentre is (A) \([(22 / 3),(10 / 3),(4 / 3)]\) (B) \([\\{(-22) / 3\\}(10 / 3),(4 / 3)]\) (C) \([(22 / 3),\\{(-10) / 3\\},(4 / 3)]\) (D) \([(22 / 3),(10 / 3),\\{(-4) / 3\\}]\)

The equation of plane passing through the point \(\mathrm{A}(1,2,3)\), \(\mathrm{B}(3,-1,2)\) also perpendicular to \(\mathrm{x}+3 \mathrm{y}+2 \mathrm{z}=7\) is (A) \(3 x+5 y-9 z+14=0\) (B) \(3 x-5 y-9 z+14=0\) (C) \(3 x-5 y+9 z+14=0\) (D) \(3 x+5 y+9 z+14=0\)

If \((\underline{a}+\underline{b}) \cdot(\underline{a}-\underline{b})=63\) and \(|\underline{a}|=8|\underline{b}|\), then \(|\underline{a}|=\) (A) 8 (B) 64 (C) 16 4

If angle between \(\underline{a}\) and \(\underline{b}\) is \(\theta\), then \([(|\underline{\mathbf{a}} \times \underline{\mathrm{b}}|) /(\underline{\mathrm{a}} \cdot \underline{\mathrm{b}})]=\) \(\begin{array}{llll}(\text { A })-\cot \theta & \text { (B) }-\tan \theta & \text { (C) } \tan \theta & \text { (D) } \cot \theta\end{array}\)
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