Question

Let two non-collinear unit vectors $\mathop a\limits^ \wedge$ and $\mathop b\limits^ \wedge$ and b form an acute angle. A point P moves so that at any time t the position vector $\mathop {OP}\limits^ \to$(where O is the origin) is given by $\mathop a\limits^ \wedge$cost +$\mathop b\limits^ \wedge$sint . When P is farthest from origin O,let M be the length of $\mathop {OP}\limits^ \to$and $\mathop u\limits^ \wedge$ be the unit vector along vector OP. Then:
A) $\mathop u\limits^ \wedge$=$\dfrac{{\mathop a\limits^ \wedge + \mathop b\limits^ \wedge }}{{\left| {\mathop a\limits^ \wedge + \mathop b\limits^ \wedge } \right|}}$ and M=${\left( {1 + \mathop a\limits^ \wedge \mathop b\limits^ \wedge } \right)^{\dfrac{1}{2}}}$
B) $\mathop u\limits^ \wedge$=$\dfrac{{\mathop a\limits^ \wedge - \mathop b\limits^ \wedge }}{{\left| {\mathop a\limits^ \wedge - \mathop b\limits^ \wedge } \right|}}$ and M=${\left( {1 + \mathop a\limits^ \wedge \mathop b\limits^ \wedge } \right)^{\dfrac{1}{2}}}$
C) $\mathop u\limits^ \wedge$=$\dfrac{{\mathop a\limits^ \wedge + \mathop b\limits^ \wedge }}{{\left| {\mathop a\limits^ \wedge + \mathop b\limits^ \wedge } \right|}}$ and M=${\left( {1 + 2\mathop a\limits^ \wedge \cdot \mathop b\limits^ \wedge } \right)^{\dfrac{1}{2}}}$
D) $\mathop u\limits^ \wedge$=$\dfrac{{\mathop a\limits^ \wedge - \mathop b\limits^ \wedge }}{{\left| {\mathop a\limits^ \wedge - \mathop b\limits^ \wedge } \right|}}$ and M=${\left( {1 + 2\mathop a\limits^ \wedge \cdot \mathop b\limits^ \wedge } \right)^{\dfrac{1}{2}}}$

Answer 1

Two non collinear unit vectors $\mathop a\limits^ \wedge$ and $\mathop b\limits^ \wedge$are given. We only need to put the position vector equation and then apply the formula for maximum value and some trigonometric function for M. After that we will get the unit vector by applying the formula of the unit vector and hence get our answer.

Complete step by step solution:
We are given some useful information in question let us write them first before starting the question:
So, we are given two non-collinear unit vectors $\mathop a\limits^ \wedge$ and $\mathop b\limits^ \wedge$ and b form an acute angle i.e. they are not in a line and have an angle which is acute.
At time t the position vector is $\mathop {OP}\limits^ \to$(where O is the origin) is given by $\mathop a\limits^ \wedge$cost +$\mathop b\limits^ \wedge$sint . Here $\mathop {OP}\limits^ \to$ is the length and maximum length of vector $\mathop {OP}\limits^ \to$ is M and $\mathop u\limits^ \wedge$ is the unit vector along vector$\mathop {OP}\limits^ \to$.
$\mathop u\limits^ \wedge$ is the unit vector along vector $\mathop {OP}\limits^ \to$ along the maximum length
Now at the position of $\mathop {OP}\limits^ \to$ vector is $\mathop a\limits^ \wedge$cost +$\mathop b\limits^ \wedge$sint ……… (1)
For getting the maximum value of (1) we will perform formula
For maximum value R=$\sqrt {{a^2} + {b^2} + 2ab}$
Now, $\mathop {OP}\limits^ \to$ can be written as, = $\sqrt {{{\left( {\mathop a\limits^ \wedge \cos t + \mathop b\limits^ \wedge \sin t} \right)}^2}}$because square and root are opposite to each other and hence no change in our original equation.
Using the above formula for maximum value in (1), $\mathop {OP}\limits^ \to$ = $\sqrt {{{\left( {\mathop a\limits^ \wedge \cos t + \mathop b\limits^ \wedge \sin t} \right)}^2}}$
On further we can write it as$\mathop {OP}\limits^ \to$ =$\sqrt {{{\left( {\mathop {a\cos t}\limits^ \wedge } \right)}^2} + {{\left( {\mathop b\limits^ \wedge \sin t} \right)}^2} + 2\mathop a\limits^ \wedge \mathop b\limits^ \wedge \sin t\cos t}$
$\Rightarrow$ $\mathop {OP}\limits^ \to$ =$\sqrt {{{\cos }^2}t + {{\sin }^2}t + \mathop a\limits^ \wedge \mathop b\limits^ \wedge \sin 2t}$
This gives $\left| {\mathop {OP}\limits^ \to } \right|$ =$\sqrt {1 + \sin 2t \cdot \mathop a\limits^ \wedge \mathop b\limits^ \wedge }$
$\therefore$ 2sintcost = sin2t
And ${\cos ^2}a + {\sin ^2}a$ =1
If we want the $\mathop {OP}\limits^ \to$vector to be maximum then sin2t should be maximum and the maximum value for sine function is 1.
Hence, sin2t is maximum at 1 or $\dfrac{\pi }{2}$
Then 2t=$\dfrac{\pi }{2}$ or t=$\dfrac{\pi }{4}$
Using all these we get final value M=${\left( {1 + \mathop a\limits^ \wedge \cdot \mathop b\limits^ \wedge } \right)^{\dfrac{1}{2}}}$ this will be farthest from the origin at t=$\dfrac{\pi }{4}$…… (2)
Putting value of t in equation (1) for the value of $\mathop {OP}\limits^ \to$
This equal to $\mathop {OP}\limits^ \to$ =$\dfrac{{\mathop a\limits^ \wedge }}{{\sqrt 2 }} + \dfrac{{\mathop b\limits^ \wedge }}{{\sqrt 2 }}$ (maximum length)…….. (3)
Taking out value of $\sqrt 2$ outside we get the unit vector of $\mathop {OP}\limits^ \to$
$\mathop {OP}\limits^ \to$ =$\dfrac{1}{{\sqrt 2 }}\dfrac{{\mathop a\limits^ \wedge + \mathop b\limits^ \wedge }}{{\left| {\dfrac{{\mathop a\limits^ \wedge + \mathop b\limits^ \wedge }}{{\sqrt 2 }}} \right|}}$ or this is a unit vector $\mathop u\limits^ \wedge$ ……. (4)
Cancelling $\sqrt 2$ and solving further equation (4) we get unit vector $\mathop u\limits^ \wedge$ =$\dfrac{{\mathop a\limits^ \wedge + \mathop b\limits^ \wedge }}{{\left| {\mathop a\limits^ \wedge + \mathop b\limits^ \wedge } \right|}}$ ………. (5)
Combining the result of (2) and (5) we get $\mathop u\limits^ \wedge$=$\dfrac{{\mathop a\limits^ \wedge + \mathop b\limits^ \wedge }}{{\left| {\mathop a\limits^ \wedge + \mathop b\limits^ \wedge } \right|}}$ and M=${\left( {1 + \mathop a\limits^ \wedge \mathop b\limits^ \wedge } \right)^{\dfrac{1}{2}}}$

Option A is the correct answer.

Note: Points to take caution:
While solving for the maximum value of M the trigonometric formula is required and also the formula for maximum value. It is advised to students to put correct value and indicate equation no so that there will be a less chance of error. While calculating the unit vector also the same thing needed.