Advertisements
Advertisements
प्रश्न
Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are `hati + 2hatj - hatk` and `-hati + hatj + hatk` respectively, externally in the ratio 2:1.
Advertisements
उत्तर
Here `veca = hati + 2hatj - hatk` and `vecb = hat-i + hatj + hatk`
The position vector of R, dividing the join of P and Q externally in the ratio 2:1 is
`vecR = (mvecb - nveca)/(m - n)`
`= (2(vecb) - 1 (veca))/(2 - 1)`
`= (2(- hati + hatj + hatk) - 1 (hati + 2hatj - hatk))/(2 - 1)`
`= -3hati + 0hatj + 3hatk`
`= -3hati + 3hatk`.
APPEARS IN
संबंधित प्रश्न
Write the position vector of the point which divides the join of points with position vectors `3veca-2vecb and 2veca+3vecb` in the ratio 2 : 1.
Find the position vector of the foot of perpendicular and the perpendicular distance from the point P with position vector
`2hati+3hatj+4hatk` to the plane `vecr` . `(2hati+hatj+3hatk)−26=0` . Also find image of P in the plane.
In Figure, identify the following vector.
Coinitial
Two vectors having the same magnitude are collinear.
Two collinear vectors having the same magnitude are equal.
If θ is the angle between two vectors `veca` and `vecb`, then `veca . vecb >= 0` only when ______.
Let `veca` and `vecb` be two unit vectors, and θ is the angle between them. Then `veca + vecb` is a unit vector if ______.
Find a vector of magnitude 4 units which is parallel to the vector \[\sqrt{3} \hat{i} + \hat{j}\]
Express \[\vec{AB}\] in terms of unit vectors \[\hat{i}\] and \[\hat{j}\], when the points are A (4, −1), B (1, 3)
Find \[\left| \vec{A} B \right|\] in each case.
Find the angle between the vectors \[\vec{a} \text{ and } \vec{b}\] where \[\vec{a} = \hat{i} - \hat{j} \text{ and } \vec{b} = \hat{j} + \hat{k}\]
Find the angle between the vectors \[\vec{a} \text{ and } \vec{b}\] \[\vec{a} = 2\hat{i} - \hat{j} + 2\hat{k} \text{ and } \vec{b} = 4\hat{i} + 4 \hat{j} - 2\hat{k}\]
Find the angle between the vectors \[\vec{a} = 2 \hat{i} - 3 \hat{j} + \hat{k} \text{ and } \vec{b} = \hat{i} + \hat{j} - 2 \hat{k}\]
Find the angle between the vectors \[\vec{a} = \hat{i} + 2 \hat{j} - \hat{k} , \vec{b} = \hat{i} - \hat{j} + \hat{k}\]
Dot product of a vector with \[\hat{i} + \hat{j} - 3\hat{k} , \hat{i} + 3\hat{j} - 2 \hat{k} \text{ and } 2 \hat{i} + \hat{j} + 4 \hat{k}\] are 0, 5 and 8 respectively. Find the vector.
If \[\vec{a,} \vec{b,} \vec{c}\] are three mutually perpendicular unit vectors, then prove that \[\left| \vec{a} + \vec{b} + \vec{c} \right| = \sqrt{3}\]
Show that the vector \[\hat{i} + \hat{j} + \hat{k}\] is equally inclined to the coordinate axes.
Show that the vectors \[\vec{a} = \frac{1}{7}\left( 2 \hat{i} + 3 \hat{j} + 6 \hat{k} \right), \vec{b} = \frac{1}{7}\left( 3\hat{i} - 6 {j} + 2 \hat{k} \right), \vec{c} = \frac{1}{7}\left( 6 \hat{i} + 2 \hat{j} - 3 {k} \right)\] are mutually perpendicular unit vectors.
For any two vectors \[\vec{a} \text{ and } \vec{b}\] show that \[\left( \vec{a} + \vec{b} \right) \cdot \left( \vec{a} - \vec{b} \right) = 0 \Leftrightarrow \left| \vec{a} \right| = \left| \vec{b} \right|\]
If \[\vec{\alpha} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k} \text{ and } \vec{\beta} = 2 \hat{i} + \hat{j} - 4 \hat{k} ,\] then express \[\vec{\beta}\] in the form of \[\vec{\beta} = \vec{\beta_1} + \vec{\beta_2} ,\] where \[\vec{\beta_1}\] is parallel to \[\vec{\alpha} \text{ and } \vec{\beta_2}\] is perpendicular to \[\vec{\alpha}\]
If either \[\vec{a} = \vec{0} \text{ or } \vec{b} = \vec{0}\] then \[\vec{a} \cdot \vec{b} = 0 .\] But the converse need not be true. Justify your answer with an example.
If A, B and C have position vectors (0, 1, 1), (3, 1, 5) and (0, 3, 3) respectively, show that ∆ ABC is right-angled at C.
Find the unit vector in the direction of vector \[\overrightarrow{PQ} ,\]
where P and Q are the points (1, 2, 3) and (4, 5, 6).
Show that the points \[A \left( 2 \hat{i} - \hat{j} + \hat{k} \right), B \left( \hat{i} - 3 \hat{j} - 5 \hat{k} \right), C \left( 3 \hat{i} - 4 \hat{j} - 4 \hat{k} \right)\] are the vertices of a right angled triangle.
If A, B, C, D are the points with position vectors `hat"i" + hat"j" - hat"k", 2hat"i" - hat"j" + 3hat"k", 2hat"i" - 3hat"k", 3hat"i" - 2hat"j" + hat"k"`, respectively, find the projection of `vec"AB"` along `vec"CD"`.
Position vector of a point P is a vector whose initial point is origin.
The unit normal to the plane 2x + y + 2z = 6 can be expressed in the vector form as
Let (h, k) be a fixed point where h > 0, k > 0. A straight line passing through this point cuts the positive direction of the coordinate axes at the points P and Q. Then the minimum area of the ΔOPQ. O being the origin, is
The altitude through vertex C of a triangle ABC, with position vectors of vertices `veca, vecb, vecc` respectively is:
If `veca, vecb, vecc` are vectors such that `[veca, vecb, vecc]` = 4, then `[veca xx vecb, vecb xx vecc, vecc xx veca]` =
Find the direction ratio and direction cosines of a line parallel to the line whose equations are 6x − 12 = 3y + 9 = 2z − 2
A line l passes through point (– 1, 3, – 2) and is perpendicular to both the lines `x/1 = y/2 = z/3` and `(x + 2)/-3 = (y - 1)/2 = (z + 1)/5`. Find the vector equation of the line l. Hence, obtain its distance from the origin.
If points A, B and C have position vectors `2hati, hatj` and `2hatk` respectively, then show that ΔABC is an isosceles triangle.
