Advertisements
Advertisements
प्रश्न
Find the vector equation of the plane passing through the points (1, 1, 1), (1, −1, 1) and (−7, −3, −5).
Advertisements
उत्तर

\[ \text{ Let } A(1, 1, 1),B(1, -1, 1) \text{ and } C(-7, -3, -5) \text{ be the coordinates } .\]
\[\text{ The required plane passes through the point} A(1, 1, 1) \text{ whose position vector is } \vec{a} = \hat{i} + \hat{j} + \hat{k} \text{ and is normal to the vector } \vec{n} \text{ given by } \]
\[ \vec{n} = \vec{AB} \times \vec{AC .} \]
\[\text{ Clearly } , \vec{AB} = \vec{OB} - \vec{OA} = \left( \hat{i} - \hat{j} + \hat{k} \right) - \left( \hat{i} + \hat{j} + \hat{k} \right) = 0 \hat{i} - 2 \hat{j} + 0 \hat{k} \]
\[ \vec{AC} = \vec{OC} - \vec{OA} = \left( - 7 \hat{i} - 3 \hat{j} - 5 \hat{k} \right) - \left( \hat{i} + \hat{j} + \hat{k} \right) = - 8 \hat{i} - 4 \hat{j} - 6 \hat{k} \]
\[ \vec{n} = \vec{AB} × \vec{AC} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\0 & - 2 & 0 \\ - 8 & - 4 & - 6\end{vmatrix} = 12 \hat{i} + 0 \hat{j} - 16 \hat{k} \]
\[\text{ The vector equation of the required plane is }\]
\[ \vec{r} . \vec{n} = \vec{a} . \vec{n} \]
\[ \Rightarrow \vec{r} . \left( 12 \hat{i} + 0 \hat{j} - 16 \hat{k} \right) = \left( \hat{i} + \hat{j} + \hat{k} \right) . \left( 12 \hat{i} + 0 \hat{j} - 16 \hat{k} \right)\]
\[ \Rightarrow \vec{r} . \left[ 4 \left( 3 \hat{i} - 4 \hat{k} \right) \right] = 12 + 0 - 16\]
\[ \Rightarrow \vec{r} . \left[ 4 \left( 3 \hat{i} - 4 \hat{k} \right) \right] = - 4\]
\[ \Rightarrow \vec{r} . \left( 3 \hat{i} - 4 \hat{k} \right) = - 1\]
\[ \Rightarrow \vec{r} . \left( 3 \hat{i} - 4 \hat{k} \right) + 1 = 0\]
APPEARS IN
संबंधित प्रश्न
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 the line passing through the point (1, 2, − 4) and perpendicular to the two lines:
`(x -8)/3 = (y+19)/(-16) = (z - 10)/7 and (x - 15)/3 = (y - 29)/8 = (z- 5)/(-5)`
Find the vector equations of the coordinate planes.
Find the vector equation of each one of following planes.
x + y − z = 5
find the equation of the plane passing through the point (1, 2, 1) and perpendicular to the line joining the points (1, 4, 2) and (2, 3, 5). Find also the perpendicular distance of the origin from this plane
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 (1, 1, −1), (6, 4, −5) and (−4, −2, 3).
Determine the value of λ for which the following planes are perpendicular to each other.
Find the equation of a plane passing through the point (−1, −1, 2) and perpendicular to the planes 3x + 2y − 3z = 1 and 5x − 4y + z = 5.
Obtain the equation of the plane passing through the point (1, −3, −2) and perpendicular to the planes x + 2y + 2z = 5 and 3x + 3y + 2z = 8.
Find the equation of the plane passing through the points (1, −1, 2) and (2, −2, 2) and which is perpendicular to the plane 6x − 2y + 2z = 9.
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 the points (2, 2, −1) and (3, 4, 2) and parallel to the line whose direction ratios are 7, 0, 6.
Find the image of the point (0, 0, 0) in the plane 3x + 4y − 6z + 1 = 0.
Find the reflection of the point (1, 2, −1) in the plane 3x − 5y + 4z = 5.
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 distance of the point (1, −2, 3) from the plane x − y + z = 5 measured along a line parallel to \[\frac{x}{2} = \frac{y}{3} = \frac{z}{- 6} .\]
Find the length and the foot of perpendicular from the point \[\left( 1, \frac{3}{2}, 2 \right)\] to the plane \[2x - 2y + 4z + 5 = 0\] .
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 general equation of a plane parallel to X-axis.
Write the distance of the plane \[\vec{r} \cdot \left( 2 \hat{i} - \hat{j} + 2 \hat{k} \right) = 12\] from the origin.
Write the equation of the plane \[\vec{r} = \vec{a} + \lambda \vec{b} + \mu \vec{c}\] in scalar product form.
Write the equation of the plane containing the lines \[\vec{r} = \vec{a} + \lambda \vec{b} \text{ and } \vec{r} = \vec{a} + \mu \vec{c} .\]
Write the position vector of the point where the line \[\vec{r} = \vec{a} + \lambda \vec{b}\] meets the plane \[\vec{r} . \vec{n} = 0 .\]
Write the intercept cut off by the plane 2x + y − z = 5 on x-axis.
Find the length of the perpendicular drawn from the origin to the plane 2x − 3y + 6z + 21 = 0.
Find the equation of the plane which bisects the line segment joining the points (−1, 2, 3) and (3, −5, 6) at right angles.
Find the vector and Cartesian equations of the plane that passes through the point (5, 2, −4) and is perpendicular to the line with direction ratios 2, 3, −1.
Find the vector equation of the plane with intercepts 3, –4 and 2 on x, y and z-axis respectively.
Find the equation of the plane through the points (2, 1, –1) and (–1, 3, 4), and perpendicular to the plane x – 2y + 4z = 10.
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`.
If the foot of perpendicular drawn from the origin to a plane is (5, – 3, – 2), then the equation of plane is `vec"r".(5hat"i" - 3hat"j" - 2hat"k")` = 38.
The point at which the normal to the curve y = `"x" + 1/"x", "x" > 0` is perpendicular to the line 3x – 4y – 7 = 0 is:
The method of splitting a single force into two perpendicular components along x-axis and y-axis is called as ______.
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
