A plane cuts intercepts 2, 2, 1 on the coordinate axes. What are the direction cosines of the normal to the plane?
Finding Direction Cosines from Plane Intercepts
The question asks for the direction cosines of the normal to a plane that cuts intercepts 2, 2, and 1 on the coordinate axes. Understanding how to find the normal vector and its direction cosines from the plane's equation is key here.
The equation of a plane in the intercept form is given by:
\(\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1\)
where \(a\), \(b\), and \(c\) are the intercepts on the x, y, and z axes, respectively.
Steps to Find Direction Cosines of the Normal
Given the intercepts are \(a=2\), \(b=2\), and \(c=1\), we can write the equation of the plane:
\(\frac{x}{2} + \frac{y}{2} + \frac{z}{1} = 1\)
To find the normal vector, we convert this equation into the general form \(Ax + By + Cz = D\). We can do this by multiplying the entire equation by the least common multiple of the denominators, which is 2:
\(2 \times \left( \frac{x}{2} + \frac{y}{2} + \frac{z}{1} \right) = 2 \times 1\)
\(x + y + 2z = 2\)
In the general equation \(Ax + By + Cz = D\), the coefficients \(A\), \(B\), and \(C\) represent the components of a normal vector to the plane. In our equation \(x + y + 2z = 2\), we have \(A=1\), \(B=1\), and \(C=2\).
So, a normal vector to the plane is \(\vec{n} = \langle 1, 1, 2 \rangle\).
Direction cosines of a vector \(\langle A, B, C \rangle\) are given by \(\left\langle \frac{A}{|\vec{n}|}, \frac{B}{|\vec{n}|}, \frac{C}{|\vec{n}|} \right\rangle\), where \(|\vec{n}|\) is the magnitude of the vector. The magnitude of the normal vector \(\vec{n} = \langle 1, 1, 2 \rangle\) is calculated as:
\(|\vec{n}| = \sqrt{A^2 + B^2 + C^2}\)
\(|\vec{n}| = \sqrt{1^2 + 1^2 + 2^2}\)
\(|\vec{n}| = \sqrt{1 + 1 + 4}\)
\(|\vec{n}| = \sqrt{6}\)
Now, we can find the direction cosines by dividing each component of the normal vector by its magnitude \(\sqrt{6}\):
- Direction cosine along x-axis: \(\frac{1}{\sqrt{6}}\)
- Direction cosine along y-axis: \(\frac{1}{\sqrt{6}}\)
- Direction cosine along z-axis: \(\frac{2}{\sqrt{6}}\)
Thus, the direction cosines of the normal to the plane are \(\left\langle \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}} \right\rangle\).
| Step | Description | Calculation/Result |
|---|---|---|
| 1 | Write plane equation in intercept form | \(\frac{x}{2} + \frac{y}{2} + \frac{z}{1} = 1\) |
| 2 | Convert to general form \(Ax+By+Cz=D\) | \(x + y + 2z = 2\) |
| 3 | Identify normal vector \(\vec{n}\) | \(\langle 1, 1, 2 \rangle\) |
| 4 | Calculate magnitude \(|\vec{n}|\) | \(\sqrt{1^2 + 1^2 + 2^2} = \sqrt{6}\) |
| 5 | Calculate direction cosines | \(\left\langle \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}} \right\rangle\) |
Revision Table: Plane and Normal Vector Concepts
| Concept | Description | Formula/Example |
|---|---|---|
| Plane Equation (Intercept Form) | Equation of a plane in terms of its intercepts \(a, b, c\) on axes. | \(\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1\) |
| Plane Equation (General Form) | Equation of a plane in a linear form. | \(Ax + By + Cz + D = 0\) or \(Ax + By + Cz = D\) |
| Normal Vector | A vector perpendicular to the plane. | From \(Ax+By+Cz=D\), \(\vec{n} = \langle A, B, C \rangle\) is a normal vector. |
| Direction Cosines | Cosines of the angles a vector makes with the positive x, y, and z axes. Represent the components of the unit vector. | For \(\vec{v}=\langle x, y, z \rangle\), DC's are \(\left\langle \frac{x}{|\vec{v}|}, \frac{y}{|\vec{v}|}, \frac{z}{|\vec{v}|} \right\rangle\). |
Additional Information: Unit Normal Vector
The direction cosines of the normal to a plane are actually the components of the unit vector normal to the plane. There are two unit normal vectors for any plane, pointing in opposite directions. If \(\vec{n}\) is a normal vector, the unit normal vectors are \(\pm \frac{\vec{n}}{|\vec{n}|}\). The direction cosines are usually specified for one of these unit vectors. In this case, \(\left\langle \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}} \right\rangle\) are the direction cosines for one of the unit normal vectors.
Similar Questions
A point on a line has coordinates (p + 1, p - 3, √2p) where p is any real number. What are the direction cosines of the line?
If L is the line with direction ratios < 3, -2, 6 > and passing through (1, -1, 1), then what are the coordinates of the points on L whose distance from (1, -1, 1) is 2 units?
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1. The direction ratios of y-axis can be <0, 4, 0>
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Which of the statements given above is/are correct?
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1. A-line in space can have infinitely many direction ratios.
2. It is possible for certain lines that the sum of the squares of direction cosines can be equal to the sum of its direction cosines.
Which of the above statements is/are correct?
If α, β and γ are the angles which the vector \(\overrightarrow {OP} \) (O being the origin) makes with the positive direction of the coordinate axes, then which of the following are correct?
1. cos 2α + cos 2β = sin 2γ
2. sin 2α + sin 2β = cos 2γ
3. sin 2α + sin 2β + sin 2γ = 2
Select the correct answer using the code given below.
Important Questions from Direction ratios and Direction cosines
A point on a line has coordinates (p + 1, p - 3, √2p) where p is any real number. What are the direction cosines of the line?
If L is the line with direction ratios < 3, -2, 6 > and passing through (1, -1, 1), then what are the coordinates of the points on L whose distance from (1, -1, 1) is 2 units?
If l, m, n are the direction cosines of the line x - 1 = 2(y + 3) = 1 - z, then what is l 4+ m 4+ n 4equal to?
Which of the following is the direction cosines of z, y and x-axis?
Determine the direction cosines of the unit vector perpendicular to the plane \(\vec{r} \cdot(2 \hat{\imath}-6 \hat{\jmath}-3 \hat{k})+1=0\) passing through the origin?
