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
591.3k+ 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, \[A = \sqrt {{a^2} + {b^2} + {c^2}} \]
Recently Updated Pages
10 examples of friction in our daily life

Write structures of the following compounds i 2 Chloro3methylpentane class 11 chemistry CBSE

What is BLO What is the full form of BLO class 8 social science CBSE

Explain the Treaty of Vienna of 1815 class 10 social science CBSE

A Paragraph on Pollution in about 100-150 Words

XIX+XXX A 49 B 51 C 55 D 44 class 5 maths CBSE

Trending doubts
Find the value of the expression given below sin 30circ class 11 maths CBSE

What do you mean by retardation What is its SI uni class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

Difference between physical and chemical change class 11 chemistry CBSE

What are derived physical quantities Give any two examples class 11 physics CBSE

Which of the following is a derived physical quantity class 11 physics CBSE

