Physics, asked by abdullahs6183, 10 months ago

If vector A = 2i + 3j - k and vector B = 4i + 6j - 2k; where i, j and ka re unit vectors. What will be angle between A and B?
(A) Ï
(B) Ï/3
(C) Ï/2
(D) 0°

Answers

Answered by amitnrw
8

Answer:

angle between vectors A & B  would be Zero

Explanation:

vector A = 2i + 3j - k

vector B = 4i + 6j - 2k

=>  vector B = 2 (2i + 3j - k)

Hence we can see they Both the vector Have just different magntude but same directions

so angle between vectors A & B  would be Zero

Answered by hukam0685
16

Explanation:

Angle between two vectors can be determined by vector dot product and vector cross product.

Vector dot product is easier to do.

 \vec A \: .\vec B  =  | \vec A|  | \vec B|  \: cos \theta \\  \\ cos \theta = \frac{ \vec A\: \vec B}{ | \vec A|  | \vec B| }   \\  \\

 \vec A = 2 \hat i + 3 \hat j - \hat k \\  \\

 |\vec A|  =  \sqrt{4 + 9 + 1}  =  \sqrt{14}  \\  \\

 \vec B = 4 \hat i + 6 \hat j - 2 \hat k \\  \\

 |\vec B|  =  \sqrt{16 + 36 + 4}  =  \sqrt{56}  \\  \\

 \vec A .\vec B  = 2 \times 4 + 6 \times 3 + 1 \times 2 \\  \\  = 8 + 18 + 2 \\  \\  = 28 \\  \\

cos \theta = \frac{ 28}{  \sqrt{14}   \sqrt{56} }   \\  \\  = \frac{ 28}{  \sqrt{14 \times 56} }  \\  \\  =  \frac{28}{28}   \\  \\cos \theta = 1 \\  \\ \theta =  {cos}^{ - 1} (1) \\  \\ \theta =  {cos}^{ - 1} (cos \: 0°) \\  \\ \theta = 0° \\  \\

Hence option 4 is correct.

Hope it helps you.

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