Question

How do you use the definition of the scalar product, find the angles between the following pairs of vectors: $- 4i + 5{\text{ }}j - k$ and $3i + 4j - k$ ?

Answer 1

Hint : For solving this particular question we have to use the definition of the scalar product to find the angle between these two given vectors. We have to calculate the dot product , Evaluate the magnitudes, and lastly substitute into the respective expression.

Complete step by step solution:
we have given pairs of vectors: $- 4i + 5{\text{ }}j - k$ and $3i + 4j - k$,
we have to use the definition of the scalar product to find the angle between these two given vectors.
Let these vectors as $\overrightarrow A$ be equal to $- 4i + 5{\text{ }}j - k$ and $\overrightarrow B$ is equal to $3i + 4j - k$ .
We can represent the given first vector as following ,
$\overrightarrow A = < \- 4,5, - 1 >$ and
Similarly, we can represent the given second vector as following ,
$\overrightarrow B = < 3,4, - 1 >$
We know that ,
The angle between the two vectors is given as
$\overrightarrow A .\overrightarrow B = \left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|\cos \theta$
Where $\overrightarrow A = < \- 4,5, - 1 >$ ,
$\overrightarrow B = < 3,4, - 1 >$ ,
$\left| {\overrightarrow A } \right|$ represents the magnitude of the vector $\overrightarrow A$ ,
$\left| {\overrightarrow B } \right|$ represents the magnitude of the vector $\overrightarrow B$ and
$\theta$ represents the angle between the two vectors that are $\overrightarrow A$ and $\overrightarrow B$.
Now, calculating the dot product ,
$\overrightarrow A .\overrightarrow B = < \- 4,5, - 1 > . < 3,4, - 1 >$
$= - 12 + 20 + 1 \\\ = 9;$
Evaluating the magnitudes ,
$\left| {\overrightarrow A } \right| = \left| { < \- 4,5, - 1 > } \right| = \sqrt {16 + 25 + 1} = \sqrt {42}$
$\left| {\overrightarrow B } \right| = \left| { < 3,4, - 1 > } \right| = \sqrt {9 + 16 + 1} = \sqrt {26}$
Lastly, substitute all the calculated value , we will get the following result ,
$\cos \theta = \dfrac{{\overrightarrow A .\overrightarrow B }}{{\left| {\overrightarrow A } \right|.\left| {\overrightarrow B } \right|}} = \dfrac{9}{{\sqrt {42} .\sqrt {26} }} = 0.27$
After simplifying we will get ,
$\cos \theta = 0.27$
Now, the angle between the two vectors is given as ,
$\theta = ar\cos (0.27) = {74.2^ \circ }$
Hence, we get the required result.
So, the correct answer is “ $\theta = ar\cos (0.27) = {74.2^ \circ }$ ”.

Note : While solving this question one must know the concept of vectors, angle between them. Questions similar in nature as that of above can be approached in a similar manner and we can solve it easily. Scalar product results in scalar and vector or cross product results in vector quantity.