हिंदी

The area of the quadrilateral ABCD, where A(0, 4, 1), B(2, 3, –1), C(4, 5, 0) and D(2, 6, 2), is equal to ______. - Mathematics

Advertisements
Advertisements

प्रश्न

The area of the quadrilateral ABCD, where A(0, 4, 1), B(2,  3, –1), C(4, 5, 0) and D(2, 6, 2), is equal to ______.

विकल्प

  • 9 sq.units

  • 18 sq.units

  • 27 sq.units

  • 81 sq.units

MCQ
रिक्त स्थान भरें

उत्तर

The area of the quadrilateral ABCD, where A(0, 4, 1), B(2,  3, –1), C(4, 5, 0) and D(2, 6, 2), is equal to 9 sq.units.

Explanation:

We have, `vec(AB) = (2 - 0)hati + (3 - 4)hatj + (-1 -1)hatk = 2hati - hatj - 2hatk`

`vec(BC) = (4 - 2)hati + (5 - 3)hatj + (0 + 1)hatk = 2hati + 2hatj + hatk`

`vec(CD) = (2 - 4)hati + (6 - 5)hatj + (2 - 0)hatk = -2hati + hatj + 2hatk`

`vec(DA) = (0 - 2)hati + (4 - 6)hatj + (1 - 2)hatk = -2hati - 2hatj - hatk`

∴ Area of quadrilateral ABCD = `|vec(AB) xx vec(BC)|`

= `|(hati, hatj, hatk),(2, -1, -2),(2, 2, 1)|`

= `|hati(-1 + 4) - hatj(2 - 4) + hatk(4 + 2)|`

= `|3hati - 6hatj + 6hatk|`

= `sqrt(9 + 36 + 36)`

= 9 sq.units

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 12: Introduction to Three Dimensional Geometry - Exercise [पृष्ठ २३८]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
अध्याय 12 Introduction to Three Dimensional Geometry
Exercise | Q 34 | पृष्ठ २३८

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

Name the octants in which the following points lie: 

(4, –3, 5)


Name the octants in which the following points lie: 

 (7, 4, –3)


Name the octants in which the following points lie: 

(–5, –4, 7) 


Find the image  of:

 (5, 2, –7) in the xy-plane.


Find the image  of: 

 (–4, 0, 0) in the xy-plane. 


Planes are drawn through the points (5, 0, 2) and (3, –2, 5) parallel to the coordinate planes. Find the lengths of the edges of the rectangular parallelepiped so formed. 


Find the distances of the point P(–4, 3, 5) from the coordinate axes. 


Find the point on y-axis which is equidistant from the points (3, 1, 2) and (5, 5, 2).


Show that the points A(3, 3, 3), B(0, 6, 3), C(1, 7, 7) and D(4, 4, 7) are the vertices of a square.


If A(–2, 2, 3) and B(13, –3, 13) are two points.
Find the locus of a point P which moves in such a way the 3PA = 2PB.


Verify the following: 

(0, 7, 10), (–1, 6, 6) and (–4, 9, –6) are vertices of a right-angled triangle.


Show that the plane ax + by cz + d = 0 divides the line joining the points (x1y1z1) and (x2y2z2) in the ratio \[- \frac{a x_1 + b y_1 + c z_1 + d}{a x_2 + b y_2 + c z_2 + d}\]


XOZ-plane divides the join of (2, 3, 1) and (6, 7, 1) in the ratio


The coordinates of the foot of the perpendicular drawn from the point P(3, 4, 5) on the yz- plane are


Find the image of the point (1, 6, 3) in the line `x/1 = (y - 1)/2 = (z - 2)/3`


The coordinates of the foot of the perpendicular drawn from the point (2, 5, 7) on the x-axis are given by ______.


Prove that the lines x = py + q, z = ry + s and x = p′y + q′, z = r′y + s′ are perpendicular if pp′ + rr′ + 1 = 0.


If the line drawn from the point (–2, – 1, – 3) meets a plane at right angle at the point (1, – 3, 3), find the equation of the plane


Find the angle between the lines whose direction cosines are given by the equations l + m + n = 0, l2 + m2 – n2 = 0


O is the origin and A is (a, b, c). Find the direction cosines of the line OA and the equation of plane through A at right angle to OA.


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 ax + by `+- (sqrt(a^2 + b^2) tan alpha)z ` = 0


Show that the points `(hati - hatj + 3hatk)` and `3(hati + hatj + hatk)` are equidistant from the plane `vecr * (5hati + 2hatj - 7hatk) + 9` = 0 and lies on opposite side of it.


The sine of the angle between the straight line `(x - 2)/3 = (y - 3)/4 = (z - 4)/5` and the plane 2x – 2y + z = 5 is ______.


The plane 2x – 3y + 6z – 11 = 0 makes an angle sin–1(α) with x-axis. The value of α is equal to ______.


The cartesian equation of the plane `vecr * (hati + hatj - hatk)` is ______.


The intercepts made by the plane 2x – 3y + 5z +4 = 0 on the co-ordinate axis are `-2, 4/3, - 4/5`.


The angle between the line `vecr = (5hati - hatj - 4hatk) + lambda(2hati - hatj + hatk)` and the plane `vec.(3hati - 4hatj - hatk)` + 5 = 0 is `sin^-1(5/(2sqrt(91)))`.


If the foot of perpendicular drawn from the origin to a plane is (5, – 3, – 2), then the equation of plane is `vecr.(5hati - 3hatj - 2hatk)` = 38.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×