Question

the angle made by the vector 12i+3j+4k with X axis​

Answer 1

Solution :

Given :-

A vector quantity is provided as

$\boxed{\bf{\vec{A}=12\hat{i}+3\hat{j}+4\hat{k}}}$

To Find :-

The angle made by given vector with x-axis.

Concept :-

This question is completely based on concept of dot product of vector quantities.

Formula of dot product is given by

$\underline{\boxed{\pink{\bf{\vec{A}\:{\tiny{\bullet}}\:\vec{B}=|\vec{A}||\vec{B}|\cos\theta}}}}$

Calculation :-

$\bigstar\bf\:|\vec{A}|=\sqrt{(12)^2+(3)^2+(4)^2}\\ \\ \bigstar\sf\:\blue{|\vec{A}|=\sqrt{169}=13\:unit}\\ \\ \implies\sf\:\vec{A}\:{\tiny{\bullet}}\:\vec{B}=|\vec{A}||\vec{B}|\cos\theta\\ \\ \implies\sf\:(12\hat{i}+3\hat{j}+4\hat{k})\:{\tiny{\bullet}}\:(\hat{i})=(13)(1)\cos\theta\\ \\ \implies\sf\:(12\times 1)(\hat{i}\times \hat{i})=13\cos\theta\\ \\ \implies\sf\:(12)(1)=13\cos\theta\\ \\ \implies\underline{\boxed{\bf{\purple{\theta=(cos)^{-1}\dfrac{12}{13}}}}}$

Additional information :-

▪ Vector quantity has both magnitude as well as direction.

Answer 2

Answer:

$\large\boxed{\sf{ { \cos }^{ - 1} ( \frac{12}{13}) }}$

Explanation:

Given a vector such that,

$\vec{A}=12\hat{i} + 3\hat{j}+4\hat{k}$

To find the angle made by the vector with X axis.

We know that, Unit vector along X axis is $\hat{i}$

Let, $\vec{B}=\hat{i}$

Also, we know that,

The dot product of two vectors is given by,

$\vec{A}.\vec{B}=|\vec{A}||\vec{B}|\cos\alpha$

Where, $\alpha$ is the angle between them.

Here, we have,

$= > |\vec{A}| = \sqrt{ {(12)}^{2} + {(3)}^{2} + {(4)}^{2} } \\ \\ = > |\vec{A}| = \sqrt{144 + 9 + 16} \\ \\ = > |\vec{A} | = \sqrt{169} \\ \\ = > |\vec{A}| = 13$

And, also, we have,

$=>|\vec{B}|=1$

Now, substitute the respective values,

Therefore, we will get,

$= > (12\hat{i} + 3\hat{j} + 4\hat{k}).(\hat{i}) = 13 \times 1 \times \cos( \alpha ) \\ \\ = > 12 = 13 \cos( \alpha ) \\ \\ = > \cos( \alpha ) = \dfrac{12}{13} \\ \\ = > \alpha = { \cos }^{ - 1} ( \frac{12}{13})$

Hence, the angle made by the vector with X axis is $\bold{ { \cos }^{ - 1} ( \frac{12}{13}) }$