Two points P and Q have cordinates (2, 3, −1) and (4, −3, 2) respectively. Find a unit vector parallel to the vector joining P with Q.
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Explanation:
|PQ|= √9+9+9
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= √27
= 3√3
Unit vector in direction of PQ: 1 magnitude of PQ x PQ
[3î +3₁ + 3k] = 3√ +3
3 3√3 = î+ 3 3 3√3 k
1 1 = Î+ + √5 +. 1 k
Thus, unit vector in direction of PQ=₁+1+ √3 1
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