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Question: Determine the angle which the vector \(A=5\widehat{i}+6\widehat{j}+5\widehat{k}\) makes with the x, ...

Determine the angle which the vector A=5i^+6j^+5k^A=5\widehat{i}+6\widehat{j}+5\widehat{k} makes with the x, y, and z axis.

Explanation

Solution

We can make use of vector algebra to find the solution. A vector is defined as a physical quantity which has both the magnitude as well as the direction. Here we are given the vector, A. It is a three-dimensional vector in x, y and z direction. It means it has its components in all the three axes.

Complete step by step answer:
The given vector is A=5i^+6j^+5k^A=5\widehat{i}+6\widehat{j}+5\widehat{k}. Let us find out the magnitude of this vector, ∣A∣=52+62+52=25+36+25=52\left| A \right|=\sqrt{{{5}^{2}}+{{6}^{2}}+{{5}^{2}}}=\sqrt{25+36+25}=5\sqrt{2}
Also, for the unit vectors i^,j^&k^\widehat{i},\widehat{j}\And \widehat{k} their magnitude is always 1.
In order to find out the angle made with each of the axis we make use of the dot product.
\Rightarrow \overrightarrow{A}.\widehat{i}=(5\widehat{i}+6\widehat{j}+5\widehat{k}).(\widehat{i)} \\\ \Rightarrow \left| A \right|\left| i \right|\cos \theta =5 \\\ \Rightarrow \left| A \right|\left| i \right|\cos \theta =5 \\\ \Rightarrow 5\sqrt{2}\cos \theta =5 \\\ \Rightarrow \cos \theta =\dfrac{1}{\sqrt{2}} \\\ \therefore \theta ={{45}^{0}} \\\
Now for y axis,
\Rightarrow \overrightarrow{A}.\widehat{j}=(5\widehat{i}+6\widehat{j}+5\widehat{k}).(\widehat{j)} \\\ \Rightarrow \left| A \right|\left| j \right|\cos \alpha =6 \\\ \Rightarrow \left| A \right|\left| j \right|\cos \alpha =6 \\\ \Rightarrow 5\sqrt{2}\cos \alpha =6 \\\ \Rightarrow \cos \alpha =\dfrac{6}{5\sqrt{2}} \\\ \Rightarrow \alpha ={{\cos }^{-1}}(\dfrac{6}{5\sqrt{2}}) \\\ \Rightarrow \alpha ={{\cos }^{-1}}(0.84) \\\ \therefore \alpha ={{32.85}^{0}} \\\
Now for z axis,
⇒A→.k^=(5i^+6j^+5k^).(k)^ ⇒∣A∣∣k∣cos⁡ζ=5 ⇒∣A∣∣k∣cos⁡ζ=5 ⇒52cos⁡ζ=5 ⇒cos⁡ζ=12 ∴ζ=450\Rightarrow \overrightarrow{A}.\widehat{k}=(5\widehat{i}+6\widehat{j}+5\widehat{k}).(\widehat{k)} \\\ \Rightarrow \left| A \right|\left| k \right|\cos \zeta =5 \\\ \Rightarrow \left| A \right|\left| k \right|\cos \zeta =5 \\\ \Rightarrow 5\sqrt{2}\cos \zeta =5 \\\ \Rightarrow \cos \zeta =\dfrac{1}{\sqrt{2}} \\\ \therefore \zeta ={{45}^{0}}

Additional Information:
There are many kinds of vectors such as zero vectors and null vectors which we get confused about many times. Most of the time we get confused that if the magnitude is zero then the given quantity cannot be termed as a vector but in actual it depends upon the state of the quantity.

Note: We should keep in mind that vectors can be multiplied with another vector in two ways- one is dot product and the other is cross product. While finding the angle usually we make use of dot product. The unit vectors are those vectors whose magnitude is 1. Also, the dot product of a vector with another vector, we just multiply the x component of one with the x component of the other, y component of one with y component of other and z component of one with z component of other.