हिंदी

A Vector Parallel to the Line of Intersection of the Planes → R ⋅ ( 3 ^ I − ^ J + ^ K ) = 1 and → R ⋅ ( ^ I + 4 ^ J − 2 ^ K ) = 2 Is(A) − 2 ^ I + 7 ^ J + 13 ^ K (B) 2 ^ I + 7 ^ J − 13 ^ K - Mathematics

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 

 

विकल्प

  •  \[- 2 \hat{i} + 7 \hat{j}+ 13 \hat{k} \]

  •   \[2 \hat{i}  + 7 \hat{j} - 13 \hat{k}\]

  •  \[-2 \hat{i}  + 7 \hat{j} + 13 \hat{k}\]

  •  \[2 \hat{i}  + 7 \hat{j} + 13 \hat{k}\]

MCQ

उत्तर

  \[2 \hat{i}  + 7 \hat{j} - 13 \hat{k}\]

\[\text{ Let the required vector be a } \hat { i }  + b \hat{j} + c \hat{ k }  . . . \left( 1 \right)\]
\[\text{ Since the vector is parallel to the line of intersection of the given planes } ,\]
\[3a - b + c = 0 . . . \left( 2 \right)\]
\[a + 4b - 2c = 0 . . . \left( 3 \right)\]
\[\text{ Solving (2) and (3), we get} \]
\[\frac{a}{- 2} = \frac{b}{7} = \frac{c}{13}\]
\[\text{ Substituting these values in (1), we get } \]
\[ - 2 \hat{i}  + 7 \hat{j}  + 13 \hat{k}  , \text{ which is the required vector } .\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 29: The Plane - MCQ [पृष्ठ ८५]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 29 The Plane
MCQ | Q 13 | पृष्ठ ८५

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 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.

 

Find the coordinates of the point where the line  \[\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{2}\]   intersects the plane x − y + z − 5 = 0. Also, find the angle between the line and the plane. 

 

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 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

\[x - y + z =\] 5  measured along the line \[x = y = z\]  . 
 

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 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.

  

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  \[\vec{r} = 3 \hat{i}  - \hat{j}  - \hat{k}  + \lambda\left( 2 \hat{i}  - 2 \hat{j} + \hat{k} \right) .\]

 

The equation of the plane through the intersection of the planes x + 2y + 3z = 4 and 2x + y − z = −5 and perpendicular to the plane 5x + 3y + 6z + 8 = 0 is



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 AB and C such that the centroid of ∆ABC is the point (abc). 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 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


The equation of the plane which cuts equal intercepts of unit length on the coordinate axes is


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).


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 line parallel to the y-axis and drawn through the point of intersection of x – 4y + 1 = 0 and 2x + y – 7 = 0.


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


The equation of the curve passing through the point `(0, pi/4)` whose differential equation is sin x cos y dx + cos x sin y dy = 0, is


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×