# If the vectors $\hat i - x\hat j - y\hat k$ and $\hat i + x\hat j + y\hat k$ are orthogonal to each other, then what is the locus of the point $\left( {x, y} \right)$?

${\text{A}}.$ A parabola

${\text{B}}.$ An ellipse

${\text{C}}.$ A circle

${\text{D}}.$ A straight line

Answer

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Hint – If two vectors are orthogonal to each other then they make an angle ${90^0}$ w.r.t. each other i.e. the dot product of these vectors should be zero, use this property to reach the answer.

Let first vector be $\vec a = \hat i - x\hat j - y\hat k$

And second vector be $\vec b = \hat i + x\hat j + y\hat k$

Now, it is given they are orthogonal to each other i.e. they make an angle ${90^0}$ w.r.t. each other.

And the dot product of these vectors should be zero.

$\therefore \left( {\vec a.\vec b} \right) = 0$

$\therefore \left( {\vec a.\vec b} \right) = \left( {\hat i - x\hat j - y\hat k} \right).\left( {\hat i + x\hat j + y\hat k} \right) = 0$

Now in the dot product of these vectors we know that the dot product of the same quantity gives us one and the dot product of different quantity gives us zero.

i.e. $\left( {\hat i.\hat i} \right) = 1,{\text{ }}\left( {\hat j.\hat j} \right){\text{ = 1, }}\left( {\hat k.\hat k} \right) = 1,{\text{ }}\left( {\hat i.\hat j} \right) = 0,{\text{ }}\left( {\hat j.\hat k} \right) = 0,{\text{ }}\left( {\hat k.\hat i} \right) = 0$, so use this property in above equation we have

$\left( {\vec a.\vec b} \right) = \left( {\hat i - x\hat j - y\hat k} \right).\left( {\hat i + x\hat j + y\hat k} \right) = 1 - {x^2} - {y^2} = 0$

$ \Rightarrow {x^2} + {y^2} = 1$

And we know this is the equation of the circle with center (0, 0) and radius 1 unit.

So the locus of point $\left( {x, y} \right)$ is a circle.

Hence option (c) is correct.

Note - In such types of questions always recall the property of orthogonal vectors which is stated above, then always remember the property of dot product which is also stated above then apply these rules and simplify, we will get the required answer which is the equation of circle.

Let first vector be $\vec a = \hat i - x\hat j - y\hat k$

And second vector be $\vec b = \hat i + x\hat j + y\hat k$

Now, it is given they are orthogonal to each other i.e. they make an angle ${90^0}$ w.r.t. each other.

And the dot product of these vectors should be zero.

$\therefore \left( {\vec a.\vec b} \right) = 0$

$\therefore \left( {\vec a.\vec b} \right) = \left( {\hat i - x\hat j - y\hat k} \right).\left( {\hat i + x\hat j + y\hat k} \right) = 0$

Now in the dot product of these vectors we know that the dot product of the same quantity gives us one and the dot product of different quantity gives us zero.

i.e. $\left( {\hat i.\hat i} \right) = 1,{\text{ }}\left( {\hat j.\hat j} \right){\text{ = 1, }}\left( {\hat k.\hat k} \right) = 1,{\text{ }}\left( {\hat i.\hat j} \right) = 0,{\text{ }}\left( {\hat j.\hat k} \right) = 0,{\text{ }}\left( {\hat k.\hat i} \right) = 0$, so use this property in above equation we have

$\left( {\vec a.\vec b} \right) = \left( {\hat i - x\hat j - y\hat k} \right).\left( {\hat i + x\hat j + y\hat k} \right) = 1 - {x^2} - {y^2} = 0$

$ \Rightarrow {x^2} + {y^2} = 1$

And we know this is the equation of the circle with center (0, 0) and radius 1 unit.

So the locus of point $\left( {x, y} \right)$ is a circle.

Hence option (c) is correct.

Note - In such types of questions always recall the property of orthogonal vectors which is stated above, then always remember the property of dot product which is also stated above then apply these rules and simplify, we will get the required answer which is the equation of circle.

Last updated date: 02nd Oct 2023

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