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प्रश्न
Find the vector equation of the plane passing through the intersection of the planes `vecr.(2hati + 2hatj - 3hatk) = 7, vecr.(2hati + 5hatj + 3hatk) = 9` and through the point (2, 1, 3)
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उत्तर
This is the vector equation of the required plane.
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संबंधित प्रश्न
Find the equation of the plane passing through the line of intersection of the planes `vecr.(hati + hatj + hatk) = 1` and `vecr.(2hati + 3hatj -hatk) + 4 = 0` and parallel to x-axis.
Find the equation of the plane which contains the line of intersection of the planes `vecrr.(hati + 2hatj + 3hatk) - 4 = 0, vecr.(2hati + htj - hatk) + 5 = 0`, and which is perpendicular to the plane `vecr.(5hati + 3hatj - 6hatk) + 8 = 0`.
Find the equation of the plane through the intersection of the planes 3x − 4y + 5z = 10 and 2x + 2y − 3z = 4 and parallel to the line x = 2y = 3z.
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Find the distance of the point P(3, 4, 4) from the point, where the line joining the points A(3, –4, –5) and B(2, –3, 1) intersects the plane 2x + y + z = 7.
Find the distance of the point (1, -5, 9) from the plane
Show that the plane whose vector equation is \[\vec{r} \cdot \left( \hat{i} + 2 \hat{j} - \hat{k} \right) = 3\] contains the line whose vector equation is \[\vec{r} = \hat{i} + \hat{j} + \lambda\left( 2 \hat{i} + \hat{j} + 4 \hat{k} \right) .\]
Find the equation of the plane determined by the intersection of the lines \[\frac{x + 3}{3} = \frac{y}{- 2} = \frac{z - 7}{6} \text{ and }\frac{x + 6}{1} = \frac{y + 5}{- 3} = \frac{z - 1}{2}\]
Find the coordinates of the point where the line \[\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{2}\] intersect the plane x − y + z − 5 = 0. Also, find the angle between the line and the plane.
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The equation of the plane through the line x + y + z + 3 = 0 = 2x − y + 3z + 1 and parallel to the line \[\frac{x}{1} = \frac{y}{2} = \frac{z}{3}\] is
A vector parallel to the line of intersection of the planes\[\vec{r} \cdot \left( 3 \hat{i} - \hat{j} + \hat{k} \right) = 1 \text{ and } \vec{r} \cdot \left( \hat{i} + 4 \hat{j} - 2 \hat{k} \right) = 2\] is
The distance of the point (−1, −5, −10) from the point of intersection of the line \[\vec{r} = 2 \hat{i}- \hat{j} + 2 \hat{k} + \lambda\left( 3 \hat{i} + 4 \hat{j}+ 12 \hat{k} \right)\] and the plane \[\vec{r} \cdot \left( \hat{i} - \hat{j} + \hat{k} \right) = 5\] is
The equation of the plane through the intersection of the planes ax + by + cz + d = 0 andlx + my + nz + p = 0 and parallel to the line y=0, z=0
Find the equation of the plane passing through the intersection of the planes `vecr . (hati + hatj + hatk)` and `vecr.(2hati + 3hatj - hatk) + 4 = 0` and parallel to the x-axis. Hence, find the distance of the plane from the x-axis.
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Find the vector equation of the line which is parallel to the vector `3hat"i" - 2hat"j" + 6hat"k"` and which passes through the point (1, –2, 3).
Show that the lines `(x - 1)/2 = (y - 2)/3 = (z - 3)/4` and `(x - 4)/5 = (y - 1)/2` = z intersect. Also, find their point of intersection.
Prove that the line through A(0, – 1, – 1) and B(4, 5, 1) intersects the line through C(3, 9, 4) and D(– 4, 4, 4).
Find the equation of the plane which is perpendicular to the plane 5x + 3y + 6z + 8 = 0 and which contains the line of intersection of the planes x + 2y + 3z – 4 = 0 and 2x + y – z + 5 = 0.
The plane ax + by = 0 is rotated about its line of intersection with the plane z = 0 through an angle α. Prove that the equation of the plane in its new position is `"a"x + "b"y +- (sqrt("a"^2 + "b"^2) tan alpha)`z = 0.
Find the equation of the plane through the intersection of the planes `vec"r" * (hat"i" + 3hat"j") - 6` = 0 and `vec"r" * (3hat"i" - hat"j" - 4hat"k")` = 0, whose perpendicular distance from origin is unity.
Find the equation of line parallel to the y-axis and drawn through the point of intersection of x – 4y + 1 = 0 and 2x + y – 7 = 0.
ABCD be a parallelogram and M be the point of intersection of the diagonals, if O is any point, then OA + OB + OC + OD is equal to
The equation of straight line through the intersection of the lines x – 2y = 1 and x + 3y = 2 and parallel to 3x + 4y = 0 is
