Advertisements
Advertisements
प्रश्न
Find the angles of a triangle whose vertices are A (0, −1, −2), B (3, 1, 4) and C (5, 7, 1).
Advertisements
उत्तर
\[ \vec{OA} = 0 \hat{i} - 1 \hat{j} - 2 \hat{k} ; \vec{OB} = 3 \hat{i} + 1 \hat{j} + 4 \hat{k} ; \vec{OC} = 5 \hat{i} + 7 \hat{j} + 1 \hat{k} \]
\[ \vec{AB} = \vec{OB} - \vec{OA} = 3 \hat{i} + 2 \hat{j} + 6 \hat{k} \Rightarrow \left| \vec{AB} \right| = \sqrt{9 + 4 + 36} = 7\]
\[ \vec{BA} = \vec{OA} - \vec{OB} = - 3 \hat{i} - 2 \hat{j} - 6 \hat{k} \Rightarrow \left| \vec{BA} \right| = \sqrt{9 + 4 + 36} = 7\]
\[ \vec{BC} = \vec{OC} - \vec{OB} = 2 \hat{i} + 6 \hat{j} - 3 \hat{k} \Rightarrow \left| \vec{BC} \right| = \sqrt{4 + 36 + 9} = 7\]
\[ \vec{CB} = \vec{OB} - \vec{OC} = - 2 \hat{i} - 6 \hat{j} + 3 \hat{k} \Rightarrow \left| \vec{CB} \right| = \sqrt{4 + 36 + 9} = 7\]
\[ \vec{CA} = \vec{OA} - \vec{OC} = - 5 \hat{i} - 8 \hat{j} - 3 \hat{k} \Rightarrow \left| \vec{CA} \right| = \sqrt{25 + 64 + 9} = \sqrt{98} = 7\sqrt{2}\]
\[ \vec{AC} = \vec{OC} - \vec{OA} = 5 \hat{i} + 8 \hat{j} + 3 \hat{k} \Rightarrow \left| \vec{AC} \right| = \sqrt{25 + 64 + 9} = \sqrt{98} = 7\sqrt{2}\]
\[\cos A = \frac{\vec{AB} . \vec{AC}}{\left| \vec{AB} \right|\left| \vec{AC} \right|} = \frac{15 + 16 + 18}{\left( 7 \right)\left( 7\sqrt{2} \right)} = \frac{49}{49\sqrt{2}} = \frac{1}{\sqrt{2}}\]
\[ \Rightarrow A = \cos^{- 1} \left( \frac{1}{\sqrt{2}} \right) = \frac{\pi}{4}\]
\[\cos B = \frac{\vec{BA} . \vec{BC}}{\left| \vec{BA} \right|\left| \vec{BC} \right|} = \frac{- 6 - 12 + 18}{\left( 7 \right)\left( 7 \right)} = \frac{0}{49} = 0\]
\[ \Rightarrow B = \cos^{- 1} \left( 0 \right) = \frac{\pi}{2}\]
\[\cos C = \frac{\vec{CB} . \vec{CA}}{\left| \vec{CB} \right|\left| \vec{CA} \right|} = \frac{10 + 48 - 9}{\left( 7 \right)\left( 7\sqrt{2} \right)} = \frac{49}{49\sqrt{2}} = \frac{1}{\sqrt{2}}\]
\[ \Rightarrow C = \cos^{- 1} \left( \frac{1}{\sqrt{2}} \right) = \frac{\pi}{4}\]
APPEARS IN
संबंधित प्रश्न
Find the direction ratios of a vector perpendicular to the two lines whose direction ratios are -2, 1, -1, and -3, -4, 1.
If `bara, barb, bar c` are the position vectors of the points A, B, C respectively and ` 2bara + 3barb - 5barc = 0` , then find the ratio in which the point C divides line segment AB.
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 position vector of a point which divides the join of points with position vectors `veca-2vecb" and "2veca+vecb`externally in the ratio 2 : 1
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.
Two collinear vectors are always equal in magnitude.
Two vectors having the same magnitude are collinear.
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.
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 ______.
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 \[\vec{a} = 2 \hat{i} - 3 \hat{j} + \hat{k} \text{ and } \vec{b} = \hat{i} + \hat{j} - 2 \hat{k}\]
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.
Dot products of a vector with vectors \[\hat{i} - \hat{j} + \hat{k} , 2\hat{ i} + \hat{j} - 3\hat{k} \text{ and } \text{i} + \hat{j} + \hat{k}\] are respectively 4, 0 and 2. Find the vector.
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 \[\vec{a} = 2 \hat{i} - \hat{j} + \hat{k}\] \[\vec{b} = \hat{i} + \hat{j} - 2 \hat{k}\] \[\vec{c} = \hat{i} + 3 \hat{j} - \hat{k}\] find λ such that \[\vec{a}\] is perpendicular to \[\lambda \vec{b} + \vec{c}\]
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 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 \[A \left( 2 \hat{i} - \hat{j} + \hat{k} \right), B \left( \hat{i} - 3 \hat{j} - 5 \hat{k} \right), C \left( 3 \hat{i} - 4 \hat{j} - 4 \hat{k} \right)\] are the vertices of a right angled triangle.
If \[\overrightarrow{AO} + \overrightarrow{OB} = \overrightarrow{BO} + \overrightarrow{OC} ,\] prove that A, B, C are collinear points.
The altitude through vertex C of a triangle ABC, with position vectors of vertices `veca, vecb, vecc` respectively is:
Assertion (A): If a line makes angles α, β, γ with positive direction of the coordinate axes, then sin2 α + sin2 β + sin2 γ = 2.
Reason (R): The sum of squares of the direction cosines of a line is 1.
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.
