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∆ class 11th non-med.
∆ Physics ques below√√√
∆ if two vectors A and B have resultant of 3i + 2j + 6k and when the two arc added one taken in opp. direction, then the resultant i + 4j - 3k. find the vectors and their magnitudes.
∆ don't spam. . bcz I'll report wrong answer.
∆ class 11th non-med.
∆ Physics ques below√√√
∆ if two vectors A and B have resultant of 3i + 2j + 6k and when the two arc added one taken in opp. direction, then the resultant i + 4j - 3k. find the vectors and their magnitudes.
∆ don't spam. . bcz I'll report wrong answer.
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Let the vectors be Axi+Ayj+Azk and Bxi+Byj+Bzk
Resultant = 3i + 2j + 6k = (Ax+Bx)i + (Ay+By)j+(Az+Bz)
= Ax+Bx = 3 .......(i)
Ay+By = 2 .......(ii)
Az+Bz = 6 .......(iii)
Now, since one vector has been reversed, it means they are being subtracted
= i + 4j - 3k = Axi+Ayj+Azk-(Bxi+Byj+Bzk)
= (Ax-Bx)i + (Ay-By)j + (Az-Bz)k
= Ax-Bx = 1 .........(iv)
Ay-By = 4 .........(v)
Az-Bz = -3 .........(vi)
Solving (i) and (iv), (ii) and (v), and (iii) and (vi) we get
Ax = 2 Bx = 1
Ay = 3 By = -1
Az = 1.5 Bz = 4.5
So, vectors = 2i+3j+1.5k and i-j+4.5k
Magnitudes = √4+9+2.25 and √1+1+20.25
= √15.25 and √22.25
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