Advertisements
Advertisements
Question
Find the direction cosines of the vector joining the points A (1, 2, -3) and B (-1, -2, 1) directed from A to B.
Solution
The given points are A (1, 2, -3) and B (-1, -2, 1).
`therefore vec(AB) = (- hati -2hatj + hatk) - (hati + 2hatj - 3hatk)`
`= -2hati - 4hatj + 4hatk`
⇒ `vec(AB) = -2hati - 4hatj + 4hatk`
`therefore |vec(AB)| = sqrt((-2)^2 + (-4)^2 + 4^2) `
`= sqrt(4 + 16 + 16) `
`= sqrt36 = 6`
∴ Hence, the direction cosines of `vec(AB)` are `((-2)/6, (-4)/6, 4/6) = ((-1)/3. (-2)/3, 2/3)`.
APPEARS IN
RELATED QUESTIONS
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.
Find the value of 'p' for which the vectors `3hati+2hatj+9hatk and hati-2phatj+3hatk` are parallel
In Figure, identify the following vector.
Coinitial
`veca and -veca` are collinear.
Find the position vector of the mid point of the vector joining the points P (2, 3, 4) and Q (4, 1, – 2).
Find the value of x for which `x(hati + hatj + hatk)` is a unit vector.
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 \[\sqrt{3} \hat{i} + \hat{j}\]
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 a unit vector parallel to the vector \[\hat{i} + \sqrt{3} \hat{j}\]
Find the angle between the vectors \[\vec{a} = 2 \hat{i} - 3 \hat{j} + \hat{k} \text{ and } \vec{b} = \hat{i} + \hat{j} - 2 \hat{k}\]
If \[\hat{a} \text{ and } \hat{b}\] are unit vectors inclined at an angle θ, prove that \[\cos\frac{\theta}{2} = \frac{1}{2}\left| \hat{a} + \hat{b} \right|\]
If \[\hat{ a } \text{ and } \hat{b }\] are unit vectors inclined at an angle θ, prove that
\[\tan\frac{\theta}{2} = \frac{\left| \hat{a} -\hat{b} \right|}{\left| \hat{a} + \hat{b} \right|}\]
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.
Find the angles of a triangle whose vertices are A (0, −1, −2), B (3, 1, 4) and C (5, 7, 1).
Find the magnitude of two vectors \[\vec{a} \text{ and } \vec{b}\] that are of the same magnitude, are inclined at 60° and whose scalar product is 1/2.
Show that the points whose position vectors are \[\vec{a} = 4 \hat{i} - 3 \hat{j} + \hat{k} , \vec{b} = 2 \hat{i} - 4 \hat{j} + 5 \hat{k} , \vec{c} = \hat{i} - \hat{j}\] form a right triangle.
Find the vector from the origin O to the centroid of the triangle whose vertices are (1, −1, 2), (2, 1, 3) and (−1, 2, −1).
Find the value of x for which \[x \left( \hat{i} + \hat{j} + \hat{k} \right)\] is a unit vector.
If \[\vec{a} = \hat{i} + \hat{j} + \hat{k} , \vec{b} = 2 \hat{i} - \hat{j} + 3 \hat{k} \text{ and }\vec{c} = \hat{i} - 2 \hat{j} + \hat{k} ,\] find a unit vector parallel to \[2 \vec{a} - \vec{b} + 3 \vec{c .}\]
If \[\overrightarrow{AO} + \overrightarrow{OB} = \overrightarrow{BO} + \overrightarrow{OC} ,\] prove that A, B, C are collinear points.
If \[\vec{a} \times \vec{b} = \vec{c} \times \vec{d} \text { and } \vec{a} \times \vec{c} = \vec{b} \times \vec{d}\] , show that \[\vec{a} - \vec{d}\] is parallel to \[\vec{b} - \vec{c}\] where \[\vec{a} \neq \vec{d} \text { and } \vec{b} \neq \vec{c}\] .
If `vec"a"` and `vec"b"` are the position vectors of A and B, respectively, find the position vector of a point C in BA produced such that BC = 1.5 BA.
Position vector of a point P is a vector whose initial point is origin.
The altitude through vertex C of a triangle ABC, with position vectors of vertices `veca, vecb, vecc` respectively is:
Area of rectangle having vertices A, B, C and D will position vector `(- hati + 1/2hatj + 4hatk), (hati + 1/2hatj + 4hatk) (hati - 1/2hatj + 4hatk)` and `(-hati - 1/2hatj + 4hatk)` is
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, internally the ratio 2:1.