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प्रश्न
The coordinates of the foot of perpendicular drawn from the origin to the plane 2x + y − 2z = 18 are ______
विकल्प
(4, 2, 4)
(−4, 2, 4)
(−4, −2, 4)
(4, 2, −4)
उत्तर
(4, 2, −4)
APPEARS IN
संबंधित प्रश्न
Find the length of the perpendicular (2, –3, 1) to the line `(x + 1)/(2) = (y - 3)/(3) = (z + 1)/(-1)`.
Find the perpendicular distance of the origin from the plane 6x – 2y + 3z – 7 = 0.
Reduce the equation `bar"r".(3hat"i" + 4hat"j" + 12hat"k")` to normal form and hence find
(i) the length of the perpendicular from the origin to the plane
(ii) direction cosines of the normal.
Find the vector equation of the plane passing through the point having position vector `hati + hatj + hatk` and perpendicular to the vector `4hati + 5hatj + 6hatk`.
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The lines `x/(1) = y/(2) = z/(3) and (x - 1)/(-2) = (y - 2)/(-4) = (z - 3)/(6)` are
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The equation of the plane passing through (2, -1, 3) and making equal intercepts on the coordinate axes is
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The direction cosines of the normal to the plane 2x – y + 2z = 3 are ______
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The equation of the plane passing through the points (1, −1, 1), (3, 2, 4) and parallel to the Y-axis is ______
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The equation of the plane in which the line `(x - 5)/(4) = (y - 7)/(4) = (z + 3)/(-5) and (x - 8)/(7) = (y - 4)/(1) = (z - 5)/(3)` lie, is
The equation of X axis is ______
If the foot of the perpendicular drawn from the origin to the plane is (4, −2, -5), then the equation of the plane is ______
Find direction cosines of the normal to the plane `bar"r"*(3hat"i" + 4hat"k")` = 5
If the normal to the plane has direction ratios 2, −1, 2 and it’s perpendicular distance from origin is 6, find its equation
Find the vector equation of the plane which bisects the segment joining A(2, 3, 6) and B(4, 3, −2) at right angles
The equation of the plane passing through the point (– 1, 2, 1) and perpendicular to the line joining the points (– 3, 1, 2) and (2, 3, 4) is ______.
Equation of the plane passing through A(-2, 2, 2), B(2, -2, -2) and perpendicular to x + 2y - 3z = 7 is ______
The equation of a plane containing the line of intersection of the planes 2x - y - 4 = 0 and y + 2z - 4 = 0 and passing through the point (1, 1, 0) is ______
The intercepts of the plane 3x - 4y + 6z = 48 on the co-ordinate axes are ______
The equation of the plane through (1, 2, -3) and (2, -2, 1) and parallel to the X-axis is ______
The equation of the plane, which bisects the line joining the points (1, 2, 3) and (3, 4, 5) at right angles is ______
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The equation of the plane passing through the points (1, –2, 1), (2, –1, –3) and (0, 1, 5) is ______.
Find the vector equation of the plane passing through the point A(–1, 2, –5) and parallel to the vectors `4hati - hatj + 3hatk` and `hati + hatj - hatk`.
If the mirror image of the point (2, 4, 7) in the plane 3x – y + 4z = 2 is (a, b, c), then 2a + b + 2c is equal to ______.
What will be the equation of plane passing through a point (1, 4, – 2) and parallel to the given plane – 2x + y – 3z = 9?
Reduce the equation `barr*(3hati - 4hatj + 12hatk)` = 3 to the normal form and hence find the length of perpendicular from the origin to the plane.
Find the vector equation of the line passing through the point (–2, 1, 4) and perpendicular to the plane `barr*(4hati - 5hatj + 7hatk)` = 15
