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Question: If \[\vec b = 3\hat i + 4\hat j\] and \[\vec a = \hat i - \hat j\], the vector having the same magni...

If b⃗=3i^+4j^\vec b = 3\hat i + 4\hat j and a⃗=i^−j^\vec a = \hat i - \hat j, the vector having the same magnitude as that of b⃗\vec b and parallel to a⃗\vec a is
A. 52(i^−j^)\dfrac{5}{{\sqrt 2 }}\left( {\hat i - \hat j} \right)
B. 52(i^+j^)\dfrac{5}{{\sqrt 2 }}\left( {\hat i + \hat j} \right)
C. 5(i^−j^)5\left( {\hat i - \hat j} \right)
D. 5(i^+j^)5\left( {\hat i + \hat j} \right)

Explanation

Solution

Determine the magnitude of vector b. The unit vector of a^\hat a is the ratio of vector a⃗\vec a and magnitude of vector a⃗\vec a. The vector c⃗\vec c will be equal to vector magnitude of b⃗\vec b and parallel to a⃗\vec a if c⃗=∣b⃗∣×unit vector of a^\vec c = \left| {\vec b} \right| \times {\text{unit vector of }}\hat a.

Complete step by step answer:
We have given two vectors b⃗=3i^+4j^\vec b = 3\hat i + 4\hat j and a⃗=i^−j^\vec a = \hat i - \hat j.Let us calculate the magnitude of vector b⃗\vec b as follows,
∣b⃗∣=32+42\left| {\vec b} \right| = \sqrt {{3^2} + {4^2}}
⇒∣b⃗∣=25\Rightarrow \left| {\vec b} \right| = \sqrt {25}
⇒∣b⃗∣=5\Rightarrow \left| {\vec b} \right| = 5 …… (1)
The unit vector of a^\hat a is the ratio of vector a⃗\vec a and magnitude of vector a⃗\vec a. Therefore,
unit vector a^=a⃗∣a⃗∣{\text{unit vector }}\hat a = \dfrac{{\vec a}}{{\left| {\vec a} \right|}}
⇒unit vector a^=i^−j^12+12\Rightarrow {\text{unit vector }}\hat a = \dfrac{{\hat i - \hat j}}{{\sqrt {{1^2} + {1^2}} }}
⇒unit vector a^=i^2−j^2\Rightarrow {\text{unit vector }}\hat a = \dfrac{{\hat i}}{{\sqrt 2 }} - \dfrac{{\hat j}}{{\sqrt 2 }} …… (2)

Let us calculate the vector c⃗\vec c of vector magnitude equal to b⃗\vec b and parallel to a⃗\vec a as,
c⃗=∣b⃗∣×unit vector of a^\vec c = \left| {\vec b} \right| \times {\text{unit vector of }}\hat a
Using equation (1) and (2) in the above equation, we get,
c⃗=5×(i^2−j^2)\vec c = 5 \times \left( {\dfrac{{\hat i}}{{\sqrt 2 }} - \dfrac{{\hat j}}{{\sqrt 2 }}} \right)
⇒c⃗=52i^−52j^\Rightarrow \vec c = \dfrac{5}{{\sqrt 2 }}\hat i - \dfrac{5}{{\sqrt 2 }}\hat j
This vector has the same magnitude as that of b⃗\vec b and it will be parallel to a⃗\vec a. To verify this, let us do the following procedure.

Let us calculate the magnitude of vector c⃗\vec c as follows,
∣c⃗∣=(52)2+(−52)2\left| {\vec c} \right| = \sqrt {{{\left( {\dfrac{5}{{\sqrt 2 }}} \right)}^2} + {{\left( { - \dfrac{5}{{\sqrt 2 }}} \right)}^2}}
⇒∣c⃗∣=(252)+(252)\Rightarrow \left| {\vec c} \right| = \sqrt {\left( {\dfrac{{25}}{2}} \right) + \left( {\dfrac{{25}}{2}} \right)}
⇒∣c⃗∣=502\Rightarrow \left| {\vec c} \right| = \sqrt {\dfrac{{50}}{2}}
⇒∣c⃗∣=5\Rightarrow \left| {\vec c} \right| = 5
Let us calculate the angle of a⃗\vec a as follows,
θ=tan⁡−1(−11)\theta = {\tan ^{ - 1}}\left( {\dfrac{{ - 1}}{1}} \right)
⇒θ=−45∘\Rightarrow \theta = - 45^\circ
Let us calculate the angle of c⃗\vec c as follows,
θ=tan⁡−1(−5252)\theta = {\tan ^{ - 1}}\left( {\dfrac{{ - \dfrac{5}{{\sqrt 2 }}}}{{\dfrac{5}{{\sqrt 2 }}}}} \right)
⇒θ=tan⁡−1(−1)\Rightarrow \theta = {\tan ^{ - 1}}\left( { - 1} \right)
∴θ=−45∘\therefore \theta = - 45^\circ

Thus, the angle made by a⃗\vec a and angle made by c⃗\vec c is the same. Therefore, these two vectors must be the same.

Note: The unit vector has a magnitude equal to 1. You can verify it in the above solution. While determining the angle made by the vector, always take the ratio of the y-component of the vector to the x-component of the vector and take the tan-inverse of the answer.