Advertisements
Advertisements
Question
The coordinates of the foot of the perpendicular drawn from the origin to a plane are (12, −4, 3). Find the equation of the plane.
Advertisements
Solution
\[\text{ The normal is passing through the points A(0, 0, 0) and B(12, -4, 3). So },\]
\[ \vec{n} = \vec{AB} = \vec{OB} - \vec{OA} =\left( \text{ 12 } \hat{i} - \text{ 4 }\hat{j} + \text{ 3 } \hat{k} \right) - \left( \text{ 0 }\hat{i} + \text{ 0 }\hat{j} + \text{ 0 } \hat{k} \right) = \text{ 12 } \hat{i} - \text{ 4 } \hat{j} + \text{ 3 } \hat{k} \]
\[ \text{ Since the plane passes through } (12, -4, 3), \vec{a} =12 \hat{i} - 4 \hat{j} + 3 \hat{k} \]
\[ \text{ We know that the vector equation of the plane passing through a point } \vec{a} \text{ and normal to } \vec{n} \text{ is} \]
\[ \vec{r} . \vec{n} = \vec{a} . \vec{n} \]
\[\text{ Substituting } \vec{a} = \hat{i} - \hat{j} + \hat{k} \text{ and } \vec{n} = \text{ 4 } \hat{i} + \text{ 2 } \hat{j} - \text{ 3 }\hat{k} , \text{ we get } \]
\[ \vec{r} . \left( \text{ 12 } \hat{i} - \text{ 4 }\hat{j} + \text{ 3 } \hat{k} \right) = \left( \text{ 12 } \hat{i} - \text{ 4 }\hat{j} + \text{ 3 }\hat{k} \right) . \left( \text{ 12 }\hat{i} - \text{ 4 }\hat{j} + \text{ 3 }\hat{k} \right)\]
\[ \Rightarrow \vec{r} . \left( \text{ 12 } \hat{i} - \text{ 4 } \hat{j} + \text{ 3 } \hat{k} \right) = 144 + 16 + 9\]
\[ \Rightarrow \vec{r} . \left( \text{ 12 } \hat{i} - \text{ 4 } \hat{j} + \text{ 3 }\hat{k} \right) = 169\]
\[ \Rightarrow \vec{r} . \left( \text{ 12 }\hat{i} - \text{ 4 } \hat{j} + \text{ 3 } \hat{k} \right) = 169\]
\[\text{ Substituting } \vec{r} = \text{ x }\hat{i} + \text{ y }\hat{j} + \text{ z }\hat{k} \text{ in the vector equation, we get } \]
\[\left( \text{ x } \hat{i} + \text{ y } \hat{j} + \text{ z }\hat{k} \right) . \left( \text{ 12 }\hat{i} - \text{ 4 }\hat{j} + \text{ 3 } \hat{k} \right) = 169\]
\[ \Rightarrow 12x - 4y + 3z = 169\]
APPEARS IN
RELATED QUESTIONS
Find the equation of the plane passing through the point (−1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.
If the points (1, 1, p) and (−3, 0, 1) be equidistant from the plane `vecr.(3hati + 4hatj - 12hatk)+ 13 = 0`, then find the value of p.
Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes `vecr = (hati - hatj + 2hatk) = 5`and `vecr.(3hati + hatj + hatk) = 6`
Find the vector equation of a plane passing through a point with position vector \[2 \hat{i} - \hat{j} + \hat{k} \] and perpendicular to the vector \[4 \hat{i} + 2 \hat{j} - 3 \hat{k} .\]
Find the Cartesian form of the equation of a plane whose vector equation is
\[\vec{r} \cdot \left( 12 \hat{i} - 3 \hat{j} + 4 \hat{k} \right) + 5 = 0\]
Find the equation of the plane that bisects the line segment joining the points (1, 2, 3) and (3, 4, 5) and is at right angle to it.
Find the vector equation of a plane which is at a distance of 3 units from the origin and has \[\hat{k}\] as the unit vector normal to it.
Find the vector equation of a plane which is at a distance of 5 units from the origin and which is normal to the vector \[\hat{i} - \text{2 } \hat{j} - \text{2 } \hat{k} .\]
Find the vector equation of the plane passing through points A (a, 0, 0), B (0, b, 0) and C(0, 0, c). Reduce it to normal form. If plane ABC is at a distance p from the origin, prove that \[\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} .\]
Find the vector equation of the plane passing through the points \[3 \hat{i} + 4 \hat{j} + 2 \hat{k} , 2 \hat{i} - 2 \hat{j} - \hat{k} \text{ and } 7 \hat{i} + 6 \hat{k} .\]
Find the equation of the plane passing through the origin and perpendicular to each of the planes x + 2y − z = 1 and 3x − 4y + z = 5.
Find the equation of the plane that contains the point (1, −1, 2) and is perpendicular to each of the planes 2x + 3y − 2z = 5 and x + 2y − 3z = 8.
Find the equation of the plane passing through (a, b, c) and parallel to the plane \[\vec{r} \cdot \left( \hat{i} + \hat{j} + \hat{k} \right) = 2 .\]
Find the vector equation of the plane through the points (2, 1, −1) and (−1, 3, 4) and perpendicular to the plane x − 2y + 4z = 10
Find the equation of the plane passing through (a, b, c) and parallel to the plane \[\vec{r} \cdot \left( \hat{i} + \hat{j} + \hat{k} \right) = 2 .\]
Find the vector equation of the line through the origin which is perpendicular to the plane \[\vec{r} \cdot \left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) = 3 .\]
Find the equation of the plane through (2, 3, −4) and (1, −1, 3) and parallel to x-axis.
Find the equation of a plane passing through the points (0, 0, 0) and (3, −1, 2) and parallel to the line \[\frac{x - 4}{1} = \frac{y + 3}{- 4} = \frac{z + 1}{7} .\]
Find the equation of the plane through the points (2, 2, −1) and (3, 4, 2) and parallel to the line whose direction ratios are 7, 0, 6.
Find the equation of the plane passing through the intersection of the planes x − 2y + z = 1 and 2x + y + z = 8 and parallel to the line with direction ratios proportional to 1, 2, 1. Also, find the perpendicular distance of (1, 1, 1) from this plane
Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane \[\vec{r} \cdot \left( \hat{i} + 2 \hat{j} - 5 \hat{k} \right) + 9 = 0 .\]
Find the coordinates of the foot of the perpendicular drawn from the point (5, 4, 2) to the line \[\frac{x + 1}{2} = \frac{y - 3}{3} = \frac{z - 1}{- 1} .\]
Hence, or otherwise, deduce the length of the perpendicular.
Find the image of the point with position vector \[3 \hat{i} + \hat{j} + 2 \hat{k} \] in the plane \[\vec{r} \cdot \left( 2 \hat{i} - \hat{j} + \hat{k} \right) = 4 .\] Also, find the position vectors of the foot of the perpendicular and the equation of the perpendicular line through \[3 \hat{i} + \hat{j} + 2 \hat{k} .\]
Find the direction cosines of the unit vector perpendicular to the plane \[\vec{r} \cdot \left( 6 \hat{i} - 3 \hat{j} - 2 \hat{k} \right) + 1 = 0\] passing through the origin.
Find the equation of the plane that contains the point (1, –1, 2) and is perpendicular to both the planes 2x + 3y – 2z = 5 and x + 2y – 3z = 8. Hence, find the distance of point P (–2, 5, 5) from the plane obtained
Find the position vector of the foot of perpendicular and the perpendicular distance from the point P with position vector \[2 \hat{i} + 3 \hat{j} + 4 \hat{k} \] to the plane \[\vec{r} . \left( 2 \hat{i} + \hat{j} + 3 \hat{k} \right) - 26 = 0\] Also find image of P in the plane.
Write the equation of the plane parallel to XOY- plane and passing through the point (2, −3, 5).
Write the equation of the plane parallel to the YOZ- plane and passing through (−4, 1, 0).
Write the equation of the plane passing through points (a, 0, 0), (0, b, 0) and (0, 0, c).
Write the equation of the plane passing through (2, −1, 1) and parallel to the plane 3x + 2y −z = 7.
Write the equation of the plane containing the lines \[\vec{r} = \vec{a} + \lambda \vec{b} \text{ and } \vec{r} = \vec{a} + \mu \vec{c} .\]
The vector equation of the plane containing the line \[\vec{r} = \left( - 2 \hat{i} - 3 \hat{j} + 4 \hat{k} \right) + \lambda\left( 3 \hat{i} - 2 \hat{j} - \hat{k} \right)\] and the point \[\hat{i} + 2 \hat{j} + 3 \hat{k} \] is
If O be the origin and the coordinates of P be (1, 2,−3), then find the equation of the plane passing through P and perpendicular to OP.
Find the vector equation of the plane which contains the line of intersection of the planes `vec("r").(hat"i"+2hat"j"+3hat"k"),-4=0, vec("r").(2hat"i"+hat"j"-hat"k")+5=0`and which is perpendicular to the plane`vec("r").(5hat"i"+3hat"j"-6hat"k"),+8=0`
The locus represented by xy + yz = 0 is ______.
The equation of a line, which is parallel to `2hat"i" + hat"j" + 3hat"k"` and which passes through the point (5, –2, 4), is `(x - 5)/2 = (y + 2)/(-1) = (z - 4)/3`.
The coordinates of the foot of the perpendicular drawn from the point A(1, 0, 3) to the join of the points B(4, 7, 1) and C(3, 5, 3) are
