Advertisements
Advertisements
प्रश्न
Find the intercepts made on the coordinate axes by the plane 2x + y − 2z = 3 and also find the direction cosines of the normal to the plane.
Advertisements
उत्तर
The given equation of the plane is 2x + y - 2z = 3
Dividng both sides by 3, we get
`(2x)/3 + y/3 +(-2z)/3 = 3/3 `
`⇒ x/((3/2)) + y/3 + z/(((-3)/2)) = 1 ... (1) `
We know that the equation of the plane whose intercepts on the coordianate axes are
`x/a + y/b +z/c = 1` .......... (2)
Comparing (1) and (2), we get
`a = 3/2 ; b =3 ; c = (-3)/2`
Finding the direction cosines of the normalThe given equation of the plane is
2x + y - 2z = 8
`⇒ (x hat(i) + y hat(j) + z hat(k) ). (2 hat(i) + hat(j) - 2 hat(k)) = 8`
`⇒ vec r . (2 hat(i) + hat(j) - 2 hat(k)) = 8 , ` which is the vector equation of the plane.(Because the vector equation of the plane is `vec r . vec n = vec a . vec n ,` where the normal to the plane `vec n = 2 vec i + vec j -2 hat(k). )`
`|vec n| = sqrt (4+1+4) =3`
So, the unit vector perpendicular to` vec n =(vec n)/(|vec n |) = ( 2 hat(i)+hat(j)-2hat(k))/3 = 2/3 hat(i) +1/3 hat(j) -2/3 hat(k) `
So, the direction cosines of the normal to the plane are, ` 2/3 , 1/3 ,(-2)/3.`
APPEARS IN
संबंधित प्रश्न
In following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
2x + 3y – z = 5
In following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
5y + 8 = 0
Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX plane.
Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the ZX − plane.
Find the coordinates of the point where the line through (3, −4, −5) and (2, − 3, 1) crosses the plane 2x + y + z = 7).
The planes: 2x − y + 4z = 5 and 5x − 2.5y + 10z = 6 are
(A) Perpendicular
(B) Parallel
(C) intersect y-axis
(C) passes through `(0,0,5/4)`
Find the coordinates of the point where the line through the points (3, - 4, - 5) and (2, - 3, 1), crosses the plane determined by the points (1, 2, 3), (4, 2,- 3) and (0, 4, 3)
Reduce the equation 2x − 3y − 6z = 14 to the normal form and, hence, find the length of the perpendicular from the origin to the plane. Also, find the direction cosines of the normal to the plane.
Write the normal form of the equation of the plane 2x − 3y + 6z + 14 = 0.
The direction ratios of the perpendicular from the origin to a plane are 12, −3, 4 and the length of the perpendicular is 5. Find the equation of the plane.
Find the equation of a plane which is at a distance of \[3\sqrt{3}\] units from the origin and the normal to which is equally inclined to the coordinate axes.
Find the vector equation of the plane which is at a distance of \[\frac{6}{\sqrt{29}}\] from the origin and its normal vector from the origin is \[2 \hat{i} - 3 \hat{j} + 4 \hat{k} .\] Also, find its Cartesian form.
Find the distance of the plane 2x − 3y + 4z − 6 = 0 from the origin.
Find the equation of the plane which contains the line of intersection of the planes \[x + 2y + 3z - 4 = 0 \text { and } 2x + y - z + 5 = 0\] and whose x-intercept is twice its z-intercept.
Find the value of λ such that the line \[\frac{x - 2}{6} = \frac{y - 1}{\lambda} = \frac{z + 5}{- 4}\] is perpendicular to the plane 3x − y − 2z = 7.
Find the equation of the plane passing through the points (−1, 2, 0), (2, 2, −1) and parallel to the line \[\frac{x - 1}{1} = \frac{2y + 1}{2} = \frac{z + 1}{- 1}\]
Find the vector equation of the plane passing through the points (3, 4, 2) and (7, 0, 6) and perpendicular to the plane 2x − 5y − 15 = 0. Also, show that the plane thus obtained contains the line \[\vec{r} = \hat{i} + 3 \hat{j} - 2 \hat{k} + \lambda\left( \hat{i} - \hat{j} + \hat{k} \right) .\]
Write the plane \[\vec{r} \cdot \left( 2 \hat{i} + 3 \hat{j} - 6 \hat{k} \right) = 14\] in normal form.
Write a vector normal to the plane \[\vec{r} = l \vec{b} + m \vec{c} .\]
Write the value of k for which the line \[\frac{x - 1}{2} = \frac{y - 1}{3} = \frac{z - 1}{k}\] is perpendicular to the normal to the plane \[\vec{r} \cdot \left( 2 \hat{i} + 3 \hat{j} + 4 \hat{k} \right) = 4 .\]
The equation of the plane containing the two lines
The equation of the plane \[\vec{r} = \hat{i} - \hat{j} + \lambda\left( \hat{i} + \hat{j} + \hat{k} \right) + \mu\left( \hat{i} - 2 \hat{j} + 3 \hat{k} \right)\] in scalar product form is
Find the image of the point having position vector `hat"i" + 3hat"j" + 4hat"k"` in the plane `hat"r" * (2hat"i" - hat"j" + hat"k") + 3` = 0.
The equations of x-axis in space are ______.
Find the equation of a plane which is at a distance `3sqrt(3)` units from origin and the normal to which is equally inclined to coordinate axis.
If the line drawn from the point (–2, – 1, – 3) meets a plane at right angle at the point (1, – 3, 3), find the equation of the plane.
The plane 2x – 3y + 6z – 11 = 0 makes an angle sin–1(α) with x-axis. The value of α is equal to ______.
The unit vector normal to the plane x + 2y +3z – 6 = 0 is `1/sqrt(14)hat"i" + 2/sqrt(14)hat"j" + 3/sqrt(14)hat"k"`.
Find the vector equation of a plane which is at a distance of 7 units from the origin and which is normal to the vector `3hati + 5hatj - 6hatk`
What will be the cartesian equation of the following plane. `vecr * (hati + hatj - hatk)` = 2
