Advertisements
Advertisements
प्रश्न
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
पर्याय
\[- 2 \hat{i} + 7 \hat{j}+ 13 \hat{k} \]
\[2 \hat{i} + 7 \hat{j} - 13 \hat{k}\]
\[-2 \hat{i} + 7 \hat{j} + 13 \hat{k}\]
\[2 \hat{i} + 7 \hat{j} + 13 \hat{k}\]
Advertisements
उत्तर
\[2 \hat{i} + 7 \hat{j} - 13 \hat{k}\]
\[\text{ Let the required vector be a } \hat { i } + b \hat{j} + c \hat{ k } . . . \left( 1 \right)\]
\[\text{ Since the vector is parallel to the line of intersection of the given planes } ,\]
\[3a - b + c = 0 . . . \left( 2 \right)\]
\[a + 4b - 2c = 0 . . . \left( 3 \right)\]
\[\text{ Solving (2) and (3), we get} \]
\[\frac{a}{- 2} = \frac{b}{7} = \frac{c}{13}\]
\[\text{ Substituting these values in (1), we get } \]
\[ - 2 \hat{i} + 7 \hat{j} + 13 \hat{k} , \text{ which is the required vector } .\]
APPEARS IN
संबंधित प्रश्न
Find the equation of the plane through the intersection of the planes 3x – y + 2z – 4 = 0 and x + y + z – 2 = 0 and the point (2, 2, 1).
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)
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.
Find the distance of the point (−1, −5, −10) from the point of intersection of the line \[\vec{r} = \left( 2 \hat{i} - \hat{j} + 2 \hat{k} \right) + \lambda\left( 3 \hat{i}+ 4 \hat{j} + 2 \hat{k} \right)\] and the plane \[\vec{r} . \left( \hat{i} - \hat{j} + \hat{k} \right) = 5 .\]
Find the distance of the point (2, 12, 5) from the point of intersection of the line \[\vec{r} = 2 \hat{i} - 4 \hat{j}+ 2 \hat{k} + \lambda\left( 3 \hat{i} + 4 \hat{j} + 2 \hat{k} \right)\] and \[\vec{r} . \left( \hat{i} - 2 \hat{j} + \hat{k} \right) = 0\]
Find the distance of the point P(−1, −5, −10) from the point of intersection of the line joining the points A(2, −1, 2) and B(5, 3, 4) with the plane \[x - y + z = 5\] .
Find the equation of the plane containing the line \[\frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z + 2}{1}\] and the point (0, 7, −7) and show that the line \[\frac{x}{1} = \frac{y - 7}{- 3} = \frac{z + 7}{2}\] also lies in the same plane.
Find the equation of the plane which contains two parallel lines\[\frac{x - 4}{1} = \frac{y - 3}{- 4} = \frac{z - 2}{5}\text{ and }\frac{x - 3}{1} = \frac{y + 2}{- 4} = \frac{z}{5} .\]
Show that the lines \[\frac{x + 4}{3} = \frac{y + 6}{5} = \frac{z - 1}{- 2}\] and 3x − 2y + z + 5 = 0 = 2x + 3y + 4z − 4 intersect. Find the equation of the plane in which they lie and also their point of intersection.
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 vector equation of the plane passing through three points with position vectors \[\hat{i} + \hat{j} - 2 \hat{k} , 2 \hat{i} - \hat{j} + \hat{k} \text{ and } \hat{i} + 2 \hat{j} + \hat{k} .\] Also, find the coordinates of the point of intersection of this plane and the line \[\vec{r} = 3 \hat{i} - \hat{j} - \hat{k} + \lambda\left( 2 \hat{i} - 2 \hat{j} + \hat{k} \right) .\]
Find the distance of the point with position vector
The plane 2x − (1 + λ) y + 3λz = 0 passes through the intersection of the planes
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 plane meets the coordinate axes at A, B and C such that the centroid of ∆ABC is the point (a, b, c). If the equation of the plane is \[\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = k,\] then k =
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 which cuts equal intercepts of unit length on the coordinate axes is
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.
Find the distance of the point (–1, –5, – 10) from the point of intersection of the line `vec"r" = 2hat"i" - hat"j" + 2hat"k" + lambda(3hat"i" + 4hat"j" + 2hat"k")` and the plane `vec"r" * (hat"i" - hat"j" + hat"k")` = 5
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.
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
