Advertisements
Advertisements
प्रश्न
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.
उत्तर
We have, A, B ,C with position vectors \[\vec{a} - 2 \vec{b} + 3 \vec{c} , 2 \vec{a} + 3 \vec{b} - 4 \vec{c} , - 7 \vec{b} + 10 \vec{c}\]
Then,
\[\overrightarrow{AB} =\] Position Vector of B - Position Vector of A
\[= 2 \vec{a} + 3 \vec{b} - 4 \vec{c} - \vec{a} + 2 \vec{b} - 3 \vec{c} \]
\[ = \vec{a} + 5 \vec{b} - 7 \vec{c}\]
\[= - 7 \vec{b} + 10 \vec{c} - 2 \vec{a} - 3 \vec{b} + 4 \vec{c} \]
\[ = - 2 \vec{a} - 10 \vec{b} + 14 \vec{c} \]
\[ = - 2 \left( \vec{a} + 5 \vec{b} - 7 \vec{c} \right)\]
∴ \[\overrightarrow{BC} = - 2 \overrightarrow{AB}\]
Hence,
But B is a point common to them.
So,
Hence, points A, B and C are collinear.
APPEARS IN
संबंधित प्रश्न
Let \[\vec{a,} \vec{b,} \vec{c,} \vec{d}\] be the position vectors of the four distinct points A, B, C, D. If \[\vec{b} - \vec{a} = \vec{c} - \vec{d}\], then show that ABCD is a parallelogram.
If \[\vec{a,} \vec{b}\] are the position vectors of A, B respectively, find the position vector of a point C in AB produced such that AC = 3 AB and that a point D in BA produced such that BD = 2BA.
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.
If the position vectors of the points A (3, 4), B (5, −6) and C (4, −1) are \[\vec{a,}\] \[\vec{b,}\] \[\vec{c}\] respectively, compute \[\vec{a} + 2 \vec{b} - 3 \vec{c}\].
If \[\vec{a}\] be the position vector whose tip is (5, −3), find the coordinates of a point B such that \[\overrightarrow{AB} =\] \[\vec{a}\], the coordinates of A being (4, −1).
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 \[\overrightarrow{PQ} = 3 \hat{i} + 2 \hat{j} - \hat{k}\] and the coordinates of P are (1, −1, 2), find the coordinates of Q.
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 a point R which divides the line segment joining points:
\[P \left( \hat{i} + 2 \hat{j} + \hat{k}\right) \text { and } Q \left( - \hat{i} + \hat{j} + \hat{k} \right)\] externally
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 \[\vec{a,} \vec{b}\] 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 \[3 \vec{a} - 2 \vec{b} + \vec{c} - 2 \vec{d} = \vec{0} .\]
Define position vector of a point.
If \[\vec{a}\], \[\vec{b}\], \[\vec{c}\] are position vectors of the points A, B and C respectively, write the value of \[\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{AC} .\]
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 with position vectors `vec"a" + vec"b"` and 2`vec"a" - vec"b"` in the ratio 1:2 is ______.
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 ______.
Position vector of the mid-point of line segment AB is `3hati + 2hatj - 3hatk`. If position vector of the point A is `2hati + 3hatj - 4hatk`, then position vector of the point B is ______.
