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Question: The coordinates of head and tail of a vector are \(\left( {2,1,0} \right)\) and \(\left( { - 4,2, - ...

The coordinates of head and tail of a vector are (2,1,0)\left( {2,1,0} \right) and (4,2,3)\left( { - 4,2, - 3} \right) respectively. Find the magnitude of the vector.
A. 23units\sqrt {23} \,{\text{units}}
B. 46units\sqrt {46} \,{\text{units}}
C. 84units\sqrt {84} \,{\text{units}}
D. 12units\sqrt {12} \,{\text{units}}

Explanation

Solution

If coordinate of a point in three-dimensional space is given as
(x,y,z)\left( {x,y,z} \right)
Then we can write the position vector of this point in vector form as
xi^+yj^+zk^x\,\hat i + y\,\hat j + z\,\hat k
If we have two vectors
P=x1i^+y1j^+z1k^\overrightarrow P = {x_1}\hat i + {y_1}\hat j + {z_1}\hat k
And
Q=x2i^+y2j^+z2k^\overrightarrow Q = {x_2}\hat i + {y_2}\hat j + {z_2}\hat k
Then
PQ\overrightarrow {PQ} is given as
PQ=(x2x1)i^+(y2y1)j^+(z2z1)k^\overrightarrow {PQ} = \left( {{x_2} - {x_1}} \right)\hat i + \left( {{y_2} - {y_1}} \right)\hat j + \left( {{z_2} - {z_1}} \right)\hat k
For a vector xi^+yj^+zk^x\,\hat i + y\,\hat j + z\,\hat kthe magnitude is given as
x2+y2+z2\sqrt {{x^2} + {y^2} + {z^2}}

Complete step by step answer:
A quantity that has both magnitude and direction is called a vector. For example, displacement, velocity etc. are vector quantities.
If coordinate of a point in three-dimensional space is given as
(x,y,z)\left( {x,y,z} \right)
Then we can write the position vector of this point in vector form as
xi^+yj^+zk^x\,\hat i + y\,\hat j + z\,\hat k
Where xx is the component in x direction, yy is the component in y direction and zz is the component in z direction. i^\hat i denotes the unit vector in x direction j^\hat j denotes the unit vector in y direction and k^\hat k denotes unit vector z direction.
Now let us express the position vector of given points in vector form.
Let head be the point A and tail be the point B
Position vector of head can be written as
A=2i^+1j^+0k^\overrightarrow A = 2\hat i + 1\hat j + 0\hat k
Position vector of tail can be written as
B=4i^+2j^+3k^\overrightarrow B = - 4\hat i + 2\hat j + - 3\hat k
Now the vector representing the line from head to tail is given by subtracting these two vectors.
Let this line be AB
If we have two vectors
P=x1i^+y1j^+z1k^\overrightarrow P = {x_1}\hat i + {y_1}\hat j + {z_1}\hat k
And
Q=x2i^+y2j^+z2k^\overrightarrow Q = {x_2}\hat i + {y_2}\hat j + {z_2}\hat k
Then
PQ\overrightarrow {PQ} is given as
PQ=(x2x1)i^+(y2y1)j^+(z2z1)k^\overrightarrow {PQ} = \left( {{x_2} - {x_1}} \right)\hat i + \left( {{y_2} - {y_1}} \right)\hat j + \left( {{z_2} - {z_1}} \right)\hat k
Using this we can write AB\overrightarrow {AB} as
AB=(42)i^+(21)j^+(30)k^ =6i^+1j^3k^  \overrightarrow {AB} = \left( { - 4 - 2} \right)\hat i + \left( {2 - 1} \right)\hat j + \left( { - 3 - 0} \right)\hat k \\\ = - 6\hat i + 1\hat j - 3\hat k \\\
For a vector xi^+yj^+zk^x\,\hat i + y\,\hat j + z\,\hat k the magnitude is given as
x2+y2+z2\sqrt {{x^2} + {y^2} + {z^2}}
Therefore, the magnitude of vector AB\overrightarrow {AB} is given as
(6)2+12+(3)2=46\sqrt {{{\left( { - 6} \right)}^2} + {1^2} + {{\left( { - 3} \right)}^2}} = \sqrt {46} units
So, the correct answer is option B.

Note: Formula to remember-
If we have two vectors
P=x1i^+y1j^+z1k^\overrightarrow P = {x_1}\hat i + {y_1}\hat j + {z_1}\hat k
And
Q=x2i^+y2j^+z2k^\overrightarrow Q = {x_2}\hat i + {y_2}\hat j + {z_2}\hat k
Then
PQ\overrightarrow {PQ} is given as
PQ=(x2x1)i^+(y2y1)j^+(z2z1)k^\overrightarrow {PQ} = \left( {{x_2} - {x_1}} \right)\hat i + \left( {{y_2} - {y_1}} \right)\hat j + \left( {{z_2} - {z_1}} \right)\hat k
For a vector xi^+yj^+zk^x\,\hat i + y\,\hat j + z\,\hat k the magnitude is given as
x2+y2+z2\sqrt {{x^2} + {y^2} + {z^2}}