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Find the position vector of a point which divides the line segment joining the points (2, −3, 4) and (3, 1, −2) externally in the ratio 3 ∶ 2. |
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Answer» Let position vector of point is \(\vec r\). The position vector of point (2, −3, 4) is \(\vec a\)= 2\(\hat i\)− 3\(\hat j\)+ 4\(\hat k\). And the position vector of point (3, 1, −2) is \(\vec b\)= 3\(\hat i\)+\(\hat j\) − 2\(\hat k\). Given that point R (x,y,z) divides the line segment joining the points A(2, −3, 4) and B(3, 1, −2) in the ratio 3 ∶ 2, externally. (By externally division formula) Therefore, the position vector of point R is \(\vec r\)= \(\frac{3\vec b - 2\vec a}{3-2}\) = 3(3\(\hat i\)+\(\hat j\) − 2\(\hat k\)) − 2(2\(\hat i\)− 3\(\hat j\) + 4\(\hat k\)) = \(\hat i\)(̂9 − 4) + \(\hat j\)(3 + 6) + \(\hat k\)(−6 − 8) = 5\(\hat i\) + 9\(\hat j\)− 14\(\hat k\). Hence, the position vector of point R is \(\vec r\) = 5\(\hat i\) + 9\(\hat j\) − 14\(\hat k\) . |
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