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Question: Given two vectors \(\vec{A}=\hat{i}-2\hat{j}-3\hat{k}\) and \(\vec{B}=4\hat{i}-2\hat{j}+6\hat{k}\), ...

Given two vectors A⃗=i^−2j^−3k^\vec{A}=\hat{i}-2\hat{j}-3\hat{k} and B⃗=4i^−2j^+6k^\vec{B}=4\hat{i}-2\hat{j}+6\hat{k}, the angle made by (A⃗+B⃗\vec{A}+\vec{B}) with the x-axis is:
(A). 30o{{30}^{o}}
(B). 45o{{45}^{o}}
(C). 60o{{60}^{o}}
(D). 90o{{90}^{o}}

Explanation

Solution

Two vectors are given; the addition of two vectors is simply the addition of coefficients of their unit vectors. The dot product of the vectors divided by the magnitude of vectors gives angle between two vectors. Substituting corresponding values in the formula, angle can be calculated. For the x-axis, we take its unit vector.

Formula used:
cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣\cos \theta =\dfrac{\vec{a}\cdot \vec{b}}{\left| {\vec{a}} \right|\left| {\vec{b}} \right|}

Complete step-by-step solution:
Vectors are used to represent physical quantities which have magnitude as well as direction. The vectors are drawn in space with a head and a tail, the head points in the direction of the quantity.
x-axis as a unit vector is represented as i^\hat{i}.

Angle between two vectors is given by-
cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣\cos \theta =\dfrac{\vec{a}\cdot \vec{b}}{\left| {\vec{a}} \right|\left| {\vec{b}} \right|} - (1)
Here, θ\theta is the angle between a⃗\vec{a} and b⃗\vec{b}
∣a⃗∣\left| {\vec{a}} \right| is the magnitude of a⃗\vec{a}
∣b⃗∣\left| {\vec{b}} \right| is the magnitude of b⃗\vec{b}

Given, A⃗=i^−2j^−3k^\vec{A}=\hat{i}-2\hat{j}-3\hat{k} and B⃗=4i^−2j^+6k^\vec{B}=4\hat{i}-2\hat{j}+6\hat{k}, then
A⃗+B⃗=5i^−4j^+3k^\vec{A}+\vec{B}=5\hat{i}-4\hat{j}+3\hat{k}

Using eq(1), the angle made by A⃗+B⃗\vec{A}+\vec{B} with the x-axis is
cos⁡θ=(5i^−4j^+3k^)⋅i^52+(−4)2+3212 ⇒cos⁡θ=552 ⇒cos⁡θ=12 ∴θ=45o \begin{aligned} & \cos \theta =\dfrac{(5\hat{i}-4\hat{j}+3\hat{k})\cdot \hat{i}}{\sqrt{{{5}^{2}}+{{(-4)}^{2}}+{{3}^{2}}}\sqrt{{{1}^{2}}}} \\\ & \Rightarrow \cos \theta =\dfrac{5}{5\sqrt{2}} \\\ & \Rightarrow \cos \theta =\dfrac{1}{\sqrt{2}} \\\ & \therefore \theta ={{45}^{o}} \\\ \end{aligned}

Therefore, the angle between A⃗+B⃗\vec{A}+\vec{B} and the x-axis is 45o{{45}^{o}}. Hence, the correct option is (B).

Additional Information:
The dot product between two vectors gives a scalar quantity. The dot product between two vectors is the product of magnitude of the vectors and the cosine of angle between them. a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ\vec{a}\cdot \vec{b}=\left| {\vec{a}} \right|\left| {\vec{b}} \right|\cos \theta . The vector product or cross product of two vectors gives a vector quantity. The vector product is the product of magnitude of the vectors and the sine of angle between them and the vector direction is denoted by a unit vector perpendicular to both vectors.a⃗×b⃗=∣a⃗∣∣b⃗∣sin⁡θn^\vec{a}\times \vec{b}=\left| {\vec{a}} \right|\left| {\vec{b}} \right|\sin \theta \hat{n}

Note:
Vectors are drawn to scale to represent the magnitude of a quantity. The other type of quantities is the scalar vectors which only have magnitude and their magnitude is represented by the unit. Vectors in space are represented in terms of three fundamental vectors i^,j^,k^\hat{i},\hat{j},\hat{k} along the x-axis, y-axis and z-axis respectively.