
The straight line $\dfrac{x-3}{3}=\dfrac{y-2}{1}=\dfrac{z-1}{0}$ is:
(a) parallel to x axis
(b) parallel to y axis
(c) parallel to z axis
(d) perpendicular to z axis
Answer
556.2k+ views
Hint: First, find the direction cosine of the given line. Then find the direction cosines of the coordinate axes. Next, find the dot product of the direction cosine of the given line to the direction cosines of the coordinate axes one by one. If any dot product comes out to be zero, then those two vectors will be perpendicular to each other.
Complete step-by-step answer:
We are given the line $\dfrac{x-3}{3}=\dfrac{y-2}{1}=\dfrac{z-1}{0}$ and we need to find its orientation with respect to the coordinate axes.
From, the equation of the line, we can see that it passes through the point (3,2,1) and it has the direction cosine, ${{a}_{1}}=$(3,1,0)
We also know the direction cosines of the coordinate axes. These are:
Direction cosines of x axis, ${{a}_{x}}=$ (1,0,0)
Direction cosines of y axis, ${{a}_{y}}=$ (0,1,0)
Direction cosines of z axis, ${{a}_{z}}=$ (0,0,1)
Now, we will take out the dot product of ${{a}_{1}}$ with ${{a}_{x}}$, ${{a}_{y}}$, and ${{a}_{z}}$ one by one.
We know that the cos product of two direction cosines: (a,b,c) and (d,e,f) is equal to $a\cdot d+b\cdot e+c\cdot f$. Using this, we will calculate the dot product of ${{a}_{1}}$ with ${{a}_{x}}$, ${{a}_{y}}$, and ${{a}_{z}}$ one by one.
${{a}_{1}}\cdot {{a}_{x}}=$ (3,1,0)$\cdot $(1,0,0) = 3 + 0 + 0 = 3
${{a}_{1}}\cdot {{a}_{y}}=$ (3,1,0)$\cdot $(0,1,0) = 0 + 1 + 0 = 1
${{a}_{1}}\cdot {{a}_{z}}=$ (3,1,0)$\cdot $(0,0,1) = 0 + 0 + 0 = 0
Now, we also know that if the dot product of two non zero vectors is zero, then they are perpendicular to each other as the dot product of two vectors is the product of their modulus and the cosine of the angle between them. Since the vectors are non zero, so the cosine should be zero which is possible only if the angle between them is ${{a}_{1}}\cdot {{a}_{z}}$ i.e. they are perpendicular to each other.
Using this fact, we find that ${{a}_{1}}\cdot {{a}_{z}}$ is equal to 0.
Thus, z axis and the given line are perpendicular.
Hence, option (d) is correct.
Note: In this question. It is very important to know the fact that if the dot product of two non-zero vectors is zero, then they are perpendicular to each other as the dot product of two vectors is the product of their modulus and the cosine of the angle between them. Since the vectors are non-zero, so the cosine should be zero which is possible only if the angle between them is ${{a}_{1}}\cdot {{a}_{z}}$ i.e. they are perpendicular to each other.
Complete step-by-step answer:
We are given the line $\dfrac{x-3}{3}=\dfrac{y-2}{1}=\dfrac{z-1}{0}$ and we need to find its orientation with respect to the coordinate axes.
From, the equation of the line, we can see that it passes through the point (3,2,1) and it has the direction cosine, ${{a}_{1}}=$(3,1,0)
We also know the direction cosines of the coordinate axes. These are:
Direction cosines of x axis, ${{a}_{x}}=$ (1,0,0)
Direction cosines of y axis, ${{a}_{y}}=$ (0,1,0)
Direction cosines of z axis, ${{a}_{z}}=$ (0,0,1)
Now, we will take out the dot product of ${{a}_{1}}$ with ${{a}_{x}}$, ${{a}_{y}}$, and ${{a}_{z}}$ one by one.
We know that the cos product of two direction cosines: (a,b,c) and (d,e,f) is equal to $a\cdot d+b\cdot e+c\cdot f$. Using this, we will calculate the dot product of ${{a}_{1}}$ with ${{a}_{x}}$, ${{a}_{y}}$, and ${{a}_{z}}$ one by one.
${{a}_{1}}\cdot {{a}_{x}}=$ (3,1,0)$\cdot $(1,0,0) = 3 + 0 + 0 = 3
${{a}_{1}}\cdot {{a}_{y}}=$ (3,1,0)$\cdot $(0,1,0) = 0 + 1 + 0 = 1
${{a}_{1}}\cdot {{a}_{z}}=$ (3,1,0)$\cdot $(0,0,1) = 0 + 0 + 0 = 0
Now, we also know that if the dot product of two non zero vectors is zero, then they are perpendicular to each other as the dot product of two vectors is the product of their modulus and the cosine of the angle between them. Since the vectors are non zero, so the cosine should be zero which is possible only if the angle between them is ${{a}_{1}}\cdot {{a}_{z}}$ i.e. they are perpendicular to each other.
Using this fact, we find that ${{a}_{1}}\cdot {{a}_{z}}$ is equal to 0.
Thus, z axis and the given line are perpendicular.
Hence, option (d) is correct.
Note: In this question. It is very important to know the fact that if the dot product of two non-zero vectors is zero, then they are perpendicular to each other as the dot product of two vectors is the product of their modulus and the cosine of the angle between them. Since the vectors are non-zero, so the cosine should be zero which is possible only if the angle between them is ${{a}_{1}}\cdot {{a}_{z}}$ i.e. they are perpendicular to each other.
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