English

The position vector of the point which divides the join of points with position vectors aba→+b→ and 2aba→-b→ in the ratio 1:2 is ______. - Mathematics

Advertisements
Advertisements

Question

The position vector of the point which divides the join of points with position vectors `vec"a" + vec"b"` and 2`vec"a" - vec"b"` in the ratio 1:2 is ______.

Options

  • `(3vec"a" + 2vec"b")/3`

  • `vec"a"`

  • `(5vec"a" - vec"b")/3`

  • `(4vec"a" + vec"b")/3`

MCQ
Fill in the Blanks
Advertisements

Solution

The position vector of the point which divides the join of points with position vectors `vec"a" + vec"b"` and 2`vec"a" - vec"b"` in the ratio 1:2 is `(4vec"a" + vec"b")/3`.

Explanation:

Applying section formula the position vector of the required point is

`(2(vec"a" + vec"b") + 1(2vec"a" - vec"b"))/(2 + 1) = (4vec"a" + vec"b")/3`

shaalaa.com
Position Vector of a Point Dividing a Line Segment in a Given Ratio
  Is there an error in this question or solution?
Chapter 10: Vector Algebra - Solved Examples [Page 212]

APPEARS IN

NCERT Exemplar Mathematics [English] Class 12
Chapter 10 Vector Algebra
Solved Examples | Q 11 | Page 212

RELATED QUESTIONS

Let \[\vec{a,} \vec{b,} \vec{c,} \vec{d}\] be the position vectors of the four distinct points ABCD. If \[\vec{b} - \vec{a} = \vec{c} - \vec{d}\], then show that ABCD is a parallelogram.


Show that the four points A, B, C, D with position vectors \[\vec{a,} \vec{b,} \vec{c,} \vec{d}\] respectively such that \[3 \vec{a} - 2 \vec{b} + 5 \vec{c} - 6 \vec{d} = 0,\] are coplanar. Also, find the position vector of the point of intersection of the line segments AC and BD.


The vertices A, B, C of triangle ABC have respectively position vectors \[\vec{a}\], \[\vec{b}\], \[\vec{c}\]  with respect to a given origin O. Show that the point D where the bisector of ∠ A meets BC has position vector \[\vec{d} = \frac{\beta \vec{b} + \gamma \vec{c}}{\beta + \gamma},\text{ where }\beta = \left| \vec{c} - \vec{a} \right| \text{ and, }\gamma = \left| \vec{a} - \vec{b} \right|\]
Hence, deduce that the incentre I has position vector
\[\frac{\alpha \vec{a} + \beta \vec{b} + \gamma \vec{c}}{\alpha + \beta + \gamma},\text{ where }\alpha = \left| \vec{b} - \vec{c} \right|\]


Prove by vector method that the internal bisectors of the angles of a triangle are concurrent.


If the position vector of a point (−4, −3) be \[\vec{a,}\] find \[\left| \vec{a} \right|\]


If the position vector \[\vec{a}\] of a point (12, n) is such that \[\left| \vec{a} \right|\] = 13, find the value (s) of n.


Show that the points 2 \[\hat{i}, -    \hat{i}-4 \] \[\hat{j}\] and \[-\hat{i}+4\hat{j}\]  form an isosceles triangle.


The position vectors of points A, B and C  are \[\lambda \hat{i} +\] 3 \[\hat{j}\],12\[\hat{i} + \mu\] \[\hat{j}\] and 11\[\hat{i} -\] 3 \[\hat{j}\] respectively. If C divides the line segment joining and B in the ratio 3:1, find the values of \[\lambda\] and \[\mu\]


Find a unit vector in the direction of the resultant of the vectors
\[\hat{i} - \hat{j} + 3 \hat{k} , 2 \hat{i} + \hat{j} - 2 \hat{k} \text{ and }\hat{i} + 2 \hat{j} - 2 \hat{k} .\]


If the vertices of a triangle are the points with position vectors \[a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k} , b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k} , c_1 \hat{i} + c_2 \hat{j} + c_3 \hat{k} ,\]
what are the vectors determined by its sides? Find the length of these vectors.


Find the position vector of a point R which divides the line segment joining points \[P \left( \hat{i} + 2 \hat{j} + \hat{k} \right) \text{ and Q }\left( - \hat{i} + \hat{j} + \hat{k} \right)\] internally 2:1.


Find the position vector of the mid-point of the vector joining the points P (2, 3, 4) and Q(4, 1, −2).


Show that the points A, B, C with position vectors \[\vec{a} - 2 \vec{b} + 3 \vec{c} , 2 \vec{a} + 3 \vec{b} - 4 \vec{c}\] and \[- 7 \vec{b} + 10 \vec{c}\] are collinear.


Prove that the points having position vectors \[\hat{i} + 2 \hat{j} + 3 \hat{k} , 3 \hat{i} + 4 \hat{j} + 7 \hat{k} , - 3 \hat{i} - 2 \hat{i} - 5 \hat{k}\] are collinear.


If the points with position vectors \[10 \hat{i} + 3 \hat{j} , 12 \hat{i} - 5 \hat{j}\text{ and a }\hat{i} + 11 \hat{j}\] are collinear, find the value of a.


If \[\vec{a,} \vec{b}\] are two non-collinear vectors prove that the points with position vectors \[\vec{a} + \vec{b,} \vec{a} - \vec{b}\text{ and }\vec{a} + \lambda \vec{b}\] are collinear for all real values of λ.


Show that the points whose position vectors are as given below are collinear:
\[2 \hat{i} + \hat{j} - \hat{k} , 3 \hat{i} - 2 \hat{j} + \hat{k} \text{ and }\hat{i} + 4 \hat{j} - 3 \hat{k}\]


Show that the four points having position vectors
\[6 \hat{i} - 7 \hat{j} , 16 \hat{i} - 19 \hat{j} - 4 \hat{k} , 3 \hat{j} - 6 \hat{k} , 2 \hat{i} - 5 \hat{j} + 10 \hat{k}\] are coplanar.


Show that the four points A, B, C and D with position vectors \[\vec{a}\], \[\vec{b}\], \[\vec{c}\], \[\vec{d}\] respectively are coplanar if and only if \[3 \vec{a} - 2 \vec{b} + \vec{c} - 2 \vec{d} = \vec{0} .\]


If \[\vec{a}\], \[\vec{b}\], \[\vec{c}\]  are position vectors of the points A, B and C respectively, write the value of \[\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{AC} .\]


If D is the mid-point of side BC of a triangle ABC such that \[\overrightarrow{AB} + \overrightarrow{AC} = \lambda \overrightarrow{AD} ,\] write the value of λ.


Find the image P' of the point P having position vector `hati+ 3hatj+ 4hatk` in the plane `vecr. (2hati - hatj + hatk) + 3 = 0 .` Hence find the length of PP'.

 

Find the position vector of the point which divides the join of points with position vectors `vec"a" + 3vec"b" and vec"a"- vec"b"` internally in the ratio 1 : 3. 


X and Y are two points with position vectors `3vec("a") + vec("b")` and `vec("a")-3vec("b")`respectively. Write the position vector of a point Z which divides the line segment XY in the ratio 2 : 1 externally.


The position vector of the point which divides the join of points `2vec"a" - 3vec"b"` and `vec"a" + vec"b"` in the ratio 3:1 is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×