
The angle between the two vectors \[\vec A = 3\hat i + 4\hat j + 5\hat k\]and\[\vec B = 3\hat i + 4\hat j - 5\hat k\] is:
A. \[{60^ \circ }\]
B. \[{0^ \circ }\]
C. \[{90^ \circ }\]
D. None of these
Answer
522k+ views
Hint:Learn about the vector operations, learn the cross product and dot product of two given vectors. To find the angle between the vectors learn how to evaluate the dot product between two given vectors.
Formula used:
The dot product of any given two vectors are given by,
\[\vec A.\vec B = AB\cos \theta \]
where, \[\vec A\] and\[\vec B\] are the two vectors, \[A\] and \[B\] are the magnitude of the vectors respectively and \[\theta \] is the angle between them.
Complete step by step answer:
We have given here two position vectors \[\vec A = 3\hat i + 4\hat j + 5\hat k\] and \[\vec B = 3\hat i + 4\hat j - 5\hat k\]. Now we have to find the angle between them. Now we know that the dot product of any two vectors is the product of their magnitude and the cosine of the angle between them. So, the dot product or scalar product of any two vectors is given by, \[\vec A.\vec B = AB\cos \theta \]
Now, here we have, \[\vec A = 3\hat i + 4\hat j + 5\hat k\] and \[\vec B = 3\hat i + 4\hat j - 5\hat k\]. So, upon calculating the dot product of these two we have,
\[\vec A.\vec B = (3\hat i + 4\hat j + 5\hat k) \cdot (3\hat i + 4\hat j - 5\hat k)\]
Now, we know that the \[\hat i\] is the unit vector along X-axis, \[\hat j\] is the unit vector along Y-axis and \[\hat k\] is the unit vector along Z-axis.
So, we have, \[\vec A.\vec B = 9 + 16 - 25 = 0\]. [since the dot product of the same vector gives the square of magnitude and the dot product of perpendicular vector gives zero.]
So we have, \[AB\cos \theta = 0\]
\[\cos \theta = 0\]
\[\Rightarrow \cos \theta = \cos {90^ \circ }\]
\[\therefore \theta = {90^ \circ }\]
So, the angle between them is \[{90^ \circ }\].
Hence, option C is the correct answer.
Note:Here since the dot product of the vectors becomes zero we did not have to find the magnitude of the vectors to find the angle, if the dot product between the vectors is not zero then we have to find the magnitude of the vectors. The magnitude of any vector \[\vec A = a\hat i + b\hat j + c\hat k\] is given by, \[A = \sqrt {{a^2} + {b^2} + {c^2}} \]
Formula used:
The dot product of any given two vectors are given by,
\[\vec A.\vec B = AB\cos \theta \]
where, \[\vec A\] and\[\vec B\] are the two vectors, \[A\] and \[B\] are the magnitude of the vectors respectively and \[\theta \] is the angle between them.
Complete step by step answer:
We have given here two position vectors \[\vec A = 3\hat i + 4\hat j + 5\hat k\] and \[\vec B = 3\hat i + 4\hat j - 5\hat k\]. Now we have to find the angle between them. Now we know that the dot product of any two vectors is the product of their magnitude and the cosine of the angle between them. So, the dot product or scalar product of any two vectors is given by, \[\vec A.\vec B = AB\cos \theta \]
Now, here we have, \[\vec A = 3\hat i + 4\hat j + 5\hat k\] and \[\vec B = 3\hat i + 4\hat j - 5\hat k\]. So, upon calculating the dot product of these two we have,
\[\vec A.\vec B = (3\hat i + 4\hat j + 5\hat k) \cdot (3\hat i + 4\hat j - 5\hat k)\]
Now, we know that the \[\hat i\] is the unit vector along X-axis, \[\hat j\] is the unit vector along Y-axis and \[\hat k\] is the unit vector along Z-axis.
So, we have, \[\vec A.\vec B = 9 + 16 - 25 = 0\]. [since the dot product of the same vector gives the square of magnitude and the dot product of perpendicular vector gives zero.]
So we have, \[AB\cos \theta = 0\]
\[\cos \theta = 0\]
\[\Rightarrow \cos \theta = \cos {90^ \circ }\]
\[\therefore \theta = {90^ \circ }\]
So, the angle between them is \[{90^ \circ }\].
Hence, option C is the correct answer.
Note:Here since the dot product of the vectors becomes zero we did not have to find the magnitude of the vectors to find the angle, if the dot product between the vectors is not zero then we have to find the magnitude of the vectors. The magnitude of any vector \[\vec A = a\hat i + b\hat j + c\hat k\] is given by, \[A = \sqrt {{a^2} + {b^2} + {c^2}} \]
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