Advertisements
Advertisements
प्रश्न
Show that the vector `hati + hatj + hatk` is equally inclined to the axes OX, OY, and OZ.
Advertisements
उत्तर
Let `veca = hati + hatj + hatk`
Then,
`|veca| =sqrt(1^2 + 1^2 + 1^2) = sqrt3`
Therefore, the direction cosines of `veca` are `(1/sqrt3, 1/sqrt3, 1/sqrt3)`.
Now, let α, β, and λ be the angles formed by `veca` with the positive directions of the x, y, and z axes.
Then, we have `cosalpha = 1/sqrt3, cosbeta = 1/sqrt3, coslambda = 1/sqrt3`.
Hence, the given vector is equally inclined to axes OX, OY, and OZ.
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.
If `bara, barb, barc` are position vectors of the points A, B, C respectively such that `3bara+ 5barb-8barc = 0`, find the ratio in which A divides BC.
Two collinear vectors are always equal in magnitude.
Two collinear vectors having the same magnitude are equal.
Find the direction cosines of the vector `hati + 2hatj + 3hatk`.
Find the value of x for which `x(hati + hatj + hatk)` is a unit vector.
Let `veca` and `vecb` be two unit vectors, and θ is the angle between them. Then `veca + vecb` is a unit vector if ______.
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 \[\vec{a} \text{ and } \vec{b}\] \[\vec{a} = 3\hat{i} - 2\hat{j} - 6\hat{k} \text{ and } \vec{b} = 4 \hat{i} - \hat{j} + 8 \hat{k}\]
Find the angles which the vector \[\vec{a} = \hat{i} -\hat {j} + \sqrt{2} \hat{k}\] makes with the coordinate axes.
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.
The adjacent sides of a parallelogram are represented by the vectors \[\vec{a} = \hat{i} + \hat{j} - \hat{k}\text{ and }\vec{b} = - 2 \hat{i} + \hat{j} + 2 \hat{k} .\]
Find unit vectors parallel to the diagonals of the parallelogram.
If \[\hat{ a } \text{ and } \hat{b }\] are unit vectors inclined at an angle θ, prove that
\[\tan\frac{\theta}{2} = \frac{\left| \hat{a} -\hat{b} \right|}{\left| \hat{a} + \hat{b} \right|}\]
If \[\vec{a,} \vec{b,} \vec{c}\] are three mutually perpendicular unit vectors, then prove that \[\left| \vec{a} + \vec{b} + \vec{c} \right| = \sqrt{3}\]
If \[\left| \vec{a} + \vec{b} \right| = 60, \left| \vec{a} - \vec{b} \right| = 40 \text{ and } \left| \vec{b} \right| = 46, \text{ find } \left| \vec{a} \right|\]
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.
If \[\vec{p} = 5 \hat{i} + \lambda \hat{j} - 3 \hat{k} \text{ and } \vec{q} = \hat{i} + 3 \hat{j} - 5 \hat{k} ,\] then find the value of λ, so that \[\vec{p} + \vec{q}\] and \[\vec{p} - \vec{q}\] are perpendicular vectors.
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.
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 vector from the origin O to the centroid of the triangle whose vertices are (1, −1, 2), (2, 1, 3) and (−1, 2, −1).
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 `hat"i" + hat"j" + hat"k", 2hat"i" + 5hat"j", 3hat"i" + 2 hat"j" - 3hat"k" and hat"i" - 6hat"j" - hat"k"` respectively are the position vectors A, B, C and D, then find the angle between the straight lines AB and CD. Find whether `vec"AB" and vec"CD"` are collinear or not.
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"`.
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 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.
