
If \[\hat i,{\rm{ }}\hat j,{\rm{ }}\hat k\] are the positive vectors of \[A,B,C\] and \[\mathop {AB}\limits^ \to = \mathop {CX}\limits^ \to \] then the positive vectors of\[X\] is
\[ - \hat i + {\rm{ }}\hat j + \hat k\]
\[\hat i - {\rm{ }}\hat j + \hat k\]
\[\hat i, + {\rm{ }}\hat j - \hat k\]
\[\hat i + {\rm{ }}\hat j + \hat k\]
Answer
574.8k+ views
Hint: Here we have to use the basic concept of the vectors equation to find out the value of the positive vectors of \[X\]. Firstly we will find out the value of the vectors\[\mathop {AB}\limits^ \to \] and \[\mathop {CX}\limits^ \to \]. Then we will equate them as it is given in the question that \[\mathop {AB}\limits^ \to = \mathop {CX}\limits^ \to \] to get the equation of the positive vectors of \[X\].
Complete step-by-step answer:
It is given in the question that the \[\mathop {AB}\limits^ \to \] vector is equal to the \[\mathop {CX}\limits^ \to \] vector.
We know that the vector \[\mathop {AB}\limits^ \to \] is the vector which is starting at point A and ending at point B. Similarly we can say that the vector \[\mathop {CX}\limits^ \to \] is the vector which is starting at the point C and ending at the point X.
We know that the equation of a vector is equal to the difference between the final point vector and the starting point vector. So, by this we will write the equation for the vectors \[\mathop {AB}\limits^ \to \] and \[\mathop {CX}\limits^ \to \]. Therefore, we get
It is given that \[\hat i,{\rm{ }}\hat j,{\rm{ }}\hat k\] are the positive vectors of \[A,B,C\] respectively.
Therefore, the equation of the vector \[\mathop {AB}\limits^ \to = \vec B - \vec A\] and we know the vectors of both the points. So, we get
Equation of the vector \[\mathop {AB}\limits^ \to = \vec B - \vec A = \hat j - \hat i\]
Similarly we will find the equation of the vector\[\mathop {CX}\limits^ \to \].
Therefore, equation of the vector \[\mathop {CX}\limits^ \to = \vec X - \vec C = \hat x - \hat k\]
We know that\[\mathop {AB}\limits^ \to = \mathop {CX}\limits^ \to \]. Therefore, we get
\[ \Rightarrow \hat j - \hat i = \hat x - \hat k\]
From this equation we will get the value of the positive vectors o f\[X\] i.e. vector \[\hat x\]
\[ \Rightarrow \hat x = - \hat i + \hat j + \hat k\]
Hence, \[ - \hat i + \hat j + \hat k\] is the positive vector of \[X\].
So, option A is the correct option.
Note: Here we have to note that the vector is the geometric object that has both the magnitude and the direction of an object. So while calculating the equation of a line vector we should know that it is equal to the difference between the final point vector and the starting point vector of that line. We should know that Vectors have three components i.e. x component, y component and z component and all the three components of the vectors are perpendicular to each other. Unit vector is a vector which has a magnitude of 1 unit and zero vector is a vector which has a magnitude of 0 unit.
Complete step-by-step answer:
It is given in the question that the \[\mathop {AB}\limits^ \to \] vector is equal to the \[\mathop {CX}\limits^ \to \] vector.
We know that the vector \[\mathop {AB}\limits^ \to \] is the vector which is starting at point A and ending at point B. Similarly we can say that the vector \[\mathop {CX}\limits^ \to \] is the vector which is starting at the point C and ending at the point X.
We know that the equation of a vector is equal to the difference between the final point vector and the starting point vector. So, by this we will write the equation for the vectors \[\mathop {AB}\limits^ \to \] and \[\mathop {CX}\limits^ \to \]. Therefore, we get
It is given that \[\hat i,{\rm{ }}\hat j,{\rm{ }}\hat k\] are the positive vectors of \[A,B,C\] respectively.
Therefore, the equation of the vector \[\mathop {AB}\limits^ \to = \vec B - \vec A\] and we know the vectors of both the points. So, we get
Equation of the vector \[\mathop {AB}\limits^ \to = \vec B - \vec A = \hat j - \hat i\]
Similarly we will find the equation of the vector\[\mathop {CX}\limits^ \to \].
Therefore, equation of the vector \[\mathop {CX}\limits^ \to = \vec X - \vec C = \hat x - \hat k\]
We know that\[\mathop {AB}\limits^ \to = \mathop {CX}\limits^ \to \]. Therefore, we get
\[ \Rightarrow \hat j - \hat i = \hat x - \hat k\]
From this equation we will get the value of the positive vectors o f\[X\] i.e. vector \[\hat x\]
\[ \Rightarrow \hat x = - \hat i + \hat j + \hat k\]
Hence, \[ - \hat i + \hat j + \hat k\] is the positive vector of \[X\].
So, option A is the correct option.
Note: Here we have to note that the vector is the geometric object that has both the magnitude and the direction of an object. So while calculating the equation of a line vector we should know that it is equal to the difference between the final point vector and the starting point vector of that line. We should know that Vectors have three components i.e. x component, y component and z component and all the three components of the vectors are perpendicular to each other. Unit vector is a vector which has a magnitude of 1 unit and zero vector is a vector which has a magnitude of 0 unit.
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