Question

The projection of a line segment on $x$, $y$, $z$ axes are 12, 4, 3. The length and the direction cosines of the line segments are
A.13, $< 12/13$, $4/13$, $3/13 >$
B.19, $< 12/19$, $4/19$ , 19 $>$
C.11, $< 12/11$, $4/11$, $3/11 >$
D.None of these

Answer 1

Here we need to find the length and the direction cosines of the line segments. First, we will write the line segment in the vector form and then we will use the given data to find the value of the vector. From there, we will find the value of length of length of the line segments using the formula and then we will find the value of direction cosines of the line segment.

Complete step-by-step answer:
Let the vector form of the line segment be represented as
$\overrightarrow a = {a_1}\overrightarrow {i} + {a_2}\overrightarrow j + {a_3}\overrightarrow k$ ……. $\left( 1 \right)$
Here vector along the coordinate axes are $\overrightarrow i$, $\overline j$ and $\overrightarrow k$
Given that projection of the given vector i.e. $\overrightarrow a$ on x-axis is 12, that with y-axis is 4 and that with z-axis is 3 i.e.
$\Rightarrow \overrightarrow a .\overrightarrow i = 12$
Now, we will find the dot product of $\overrightarrow a$ and $\overrightarrow i$ from equation 1.
$\overrightarrow a \cdot \overrightarrow i = {a_1}\overrightarrow {i} \cdot \overrightarrow i + {a_2}\overrightarrow j \cdot \overrightarrow i + {a_3}\overrightarrow k \cdot \overrightarrow i$
We know the value of dot products $\overrightarrow j \cdot \overrightarrow {i}$ and $\overrightarrow k \cdot \overrightarrow i$ is equal to zero and the value of dot product $\overrightarrow i \cdot \overrightarrow i$ is equal to 1.
Now, we will substitute the values here.
$\Rightarrow \overrightarrow a \cdot \overrightarrow i = {a_1} = 12$
It is also given that
$\overrightarrow a \cdot \overrightarrow j = 4$
Now, we will find the dot product of $\overrightarrow a$ and $\overrightarrow j$ from equation 1.
$\overrightarrow a \cdot \overrightarrow j = {a_1}\overrightarrow {i} \cdot \overrightarrow j + {a_2}\overrightarrow j \cdot \overrightarrow j + {a_3}\overrightarrow k \cdot \overrightarrow j$
We know the value of dot products $\overrightarrow {i} \cdot \overrightarrow j$ and $\overrightarrow k \cdot \overrightarrow j$ is equal to zero and the value of dot product $\overrightarrow j \cdot \overrightarrow j$ is equal to 1.
Now, we will substitute the values here.
$\Rightarrow \overrightarrow a \cdot \overrightarrow j = {a_2} = 4$
It is also given that
$\overrightarrow a \cdot \overrightarrow k = 3$
Now, we will find the dot product of $\overrightarrow a$ and $\overrightarrow k$ from equation 1.
$\Rightarrow \overrightarrow a \cdot \overrightarrow k = {a_1}\overrightarrow {i} \cdot \overrightarrow k + {a_2}\overrightarrow j \cdot \overrightarrow k + {a_3}\overrightarrow k \cdot \overrightarrow k$
We know the value of dot products $\overrightarrow {i} \cdot \overrightarrow k$ and $\overrightarrow j \cdot \overrightarrow k$ is equal to zero and the value of dot product $\overrightarrow k \cdot \overrightarrow k$ is equal to 1.
Now, we will substitute the values here.
$\Rightarrow \overrightarrow a \cdot \overrightarrow k = {a_3} = 3$
Therefore,
$\overrightarrow a = 12\overrightarrow {i} + 4\overrightarrow j + 3\overrightarrow k$
Now, we will find the length of the line segment by find the magnitude of the given vector $\overrightarrow a$.
$\Rightarrow \left| {\overrightarrow a } \right| = \sqrt {{a_1}^2 + {a_3}^2 + {a_3}^2}$
Now, we will substitute the values here.
$\Rightarrow \left| {\overrightarrow a } \right| = \sqrt {{{12}^2} + {4^2} + {3^2}}$
On further simplification, we get
$\Rightarrow \left| {\overrightarrow a } \right| = \sqrt {144 + 16 + 9}$
On adding the numbers, we get
$\Rightarrow \left| {\overrightarrow a } \right| = \sqrt {169}$
Putting the value of square root of 169, we get
$\Rightarrow \left| {\overrightarrow a } \right| = 13$
Thus, the length of the line segment is equal to 13.
Now, we will find the direction cosines of the vector.
Let the directional cosines of vector $a$ be $cos\alpha$,$cos\beta$ and $cos\gamma$.
We have
$\Rightarrow cos\alpha = \dfrac{{\overrightarrow a \cdot \overrightarrow i }}{{\mid \overrightarrow a \mid }}$
On substituting the values of vectors and their magnitude here, we get
$\Rightarrow cos\alpha = \dfrac{{12}}{{13}}$
We have
$\Rightarrow cos\beta = \dfrac{{\overrightarrow a \cdot \overrightarrow j }}{{\mid \overrightarrow a \mid }}$
On substituting the values of vectors and their magnitude here, we get
$\Rightarrow cos\beta = \dfrac{{4}}{{13}}$
We have
$\Rightarrow cos\gamma = \dfrac{{\overrightarrow a \cdot \overrightarrow k }}{{\mid \overrightarrow a \mid }}$
On substituting the values of vectors and their magnitude here, we get
$\Rightarrow cos\gamma = \dfrac{{3}}{{13}}$
Thus, the direction cosines of the vector $a$ are $\dfrac{{12}}{{13}}$, $\dfrac{4}{{13}}$ and $\dfrac{3}{{13}}$
Hence, the correct option is option A.

Note: Here, we have obtained the direction ratios of the vector. The directional cosines of a vector are defined as the cosines of the angle that a vector makes with the three coordinate axes. So, we can say direction cosine is a set of information that represents the direction of a vector. A vector has both magnitude and direction.