Advertisements
Advertisements
प्रश्न
Show that the vector \[\hat{i} + \hat{j} + \hat{k}\] is equally inclined to the coordinate axes.
Advertisements
उत्तर
\[\text{ Let } \theta_1 \text{ be the angle between } \vec{a} \text{ and } x - axis.\]
\[\left| \vec{a} \right| = \sqrt{\left( 1 \right)^2 + \left( 1 \right)^2 + \left( 1 \right)^2} = \sqrt{3}\]
\[ \vec{b} = \hat{i} .........................\text{ (Because } \ \hat{i} \text{ is the unit vector along }x-axis)\]
\[\left| \vec{b} \right| = \sqrt{\left( 1 \right)^2} = \sqrt{1} = 1\]
\[ \vec{a} . \vec{b} = 1 + 0 + 0 = 1\]
\[\cos \theta_1 = \frac{\vec{a} . \vec{b}}{\left| \vec{a} \right| \left| \vec{b} \right|} = \frac{1}{\left( \sqrt{3} \right)\left( 1 \right)} = \frac{1}{\sqrt{3}}\]
\[ \Rightarrow \theta_1 = \cos^{- 1} \left( \frac{1}{\sqrt{3}} \right) . . . \left( 1 \right)\]
\[\]
\[\text{ Let } \theta_2 \text{ be the angle between } \vec{a} \ \text{ and } y - \text{ axis }\]
\[\left| \vec{a} \right| = \sqrt{\left( 1 \right)^2 + \left( 1 \right)^2 + \left( 1 \right)^2} = \sqrt{3}\]
\[ \vec{b} = \hat{j} ........................... \text { (Because } \hat{j}\ \text{ is the unit vector along }y-\text{ axis) }\]
\[\left| \vec{b} \right| = \sqrt{\left( 1 \right)^2} = \sqrt{1} = 1\]
\[ \vec{a} . \vec{b} = 0 + 1 + 0 = 1\]
\[\cos \theta_2 = \frac{\vec{a} . \vec{b}}{\left| \vec{a} \right| \left| \vec{b} \right|} = \frac{1}{\left( \sqrt{3} \right)\left( 1 \right)} = \frac{1}{\sqrt{3}}\]
\[ \Rightarrow \theta_2 = \cos^{- 1} \left( \frac{1}{\sqrt{3}} \right) . . . \left( 2 \right)\]
\[\text{ Let } \theta_3 \text{ be the angle between } \vec{a} \text{ and } z - \text{ axis }.\]
\[\left| \vec{a} \right| = \sqrt{\left( 1 \right)^2 + \left( 1 \right)^2 + \left( 1 \right)^2} = \sqrt{3}\]
\[ \vec{b} = \hat{k}.................................. { (Because } \ \hat{k}\ \text{ is the unit vector along }z-\text{ axis })\]
\[\left| \vec{b} \right| = \sqrt{\left( 1 \right)^2} = \sqrt{1} = 1\]
\[ \vec{a} . \vec{b} = 0 + 0 + 1 = 1\]
\[\cos \theta = \frac{\vec{a} . \vec{b}}{\left| \vec{a} \right| \left| \vec{b} \right|} = \frac{1}{\left( \sqrt{3} \right)\left( 1 \right)} = \frac{1}{\sqrt{3}}\]
\[ \Rightarrow \theta = \cos^{- 1} \left( \frac{1}{\sqrt{3}} \right) . . . \left( 3 \right)\]
\[\text{ From } (1), (2) \text{ and } (3), \text{ the given vector is equally inclined to the coordinate axes }.\]
APPEARS IN
संबंधित प्रश्न
Find the value of 'p' for which the vectors `3hati+2hatj+9hatk and hati-2phatj+3hatk` are parallel
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.
Classify the following as scalar and vector quantity.
Time period
In Figure, identify the following vector.
Coinitial
`veca and -veca` are collinear.
Two collinear vectors are always equal in magnitude.
Find the direction cosines of the vector `hati + 2hatj + 3hatk`.
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.
Show that the points A, B and C with position vectors `veca = 3hati - 4hatj - 4hatk`, `vecb = 2hati - hatj + hatk` and `vecc = hati - 3hatj - 5hatk`, respectively form the vertices of a right angled triangle.
Find the value of x for which `x(hati + hatj + hatk)` is a unit vector.
ABCD is a parallelogram. If the coordinates of A, B, C are (−2, −1), (3, 0) and (1, −2) respectively, find the coordinates of D.
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 \[\vec{a} = \hat{i} + 2 \hat{j} - \hat{k} , \vec{b} = \hat{i} - \hat{j} + \hat{k}\]
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 \[\vec{a,} \vec{b,} \vec{c}\] 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.
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 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).
If the vertices A, B and C of ∆ABC have position vectors (1, 2, 3), (−1, 0, 0) and (0, 1, 2), respectively, what is the magnitude of ∠ABC?
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 \[\vec{a} \times \vec{b} = \vec{c} \times \vec{d} \text { and } \vec{a} \times \vec{c} = \vec{b} \times \vec{d}\] , show that \[\vec{a} - \vec{d}\] is parallel to \[\vec{b} - \vec{c}\] where \[\vec{a} \neq \vec{d} \text { and } \vec{b} \neq \vec{c}\] .
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:
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
If points A, B and C have position vectors `2hati, hatj` and `2hatk` respectively, then show that ΔABC is an isosceles triangle.
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.
