मराठी

Show that the Points a ( 2 ^ I − ^ J + ^ K ) , B ( ^ I − 3 ^ J − 5 ^ K ) , C ( 3 ^ I − 4 ^ J − 4 ^ K ) Are the Vertices of a Right Angled Triangle. - Mathematics

Advertisements
Advertisements

प्रश्न

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.

बेरीज
Advertisements

उत्तर

Given the points \[A\left( 2 \hat{i} - \hat{j} + \hat{k} \right), B\left( \hat{i} - 3 \hat{j} - 5 \hat{k} \right)\] and \[C\left( 3 \hat{i} - 4 \hat{j} - 4 \hat{k} \right) .\] Then, \[\vec{AB} =\] Position vector of B - Position vector of A 
\[= \hat{i} - 3 \hat{j} - 5 \hat{k} - \left( 2 \hat{i} - \hat{j} + \hat{k} \right)\]
\[ = \hat{i} - 3 \hat{j} - 5 \hat{k} - 2 \hat{i} + \hat{j} - \hat{k} \]
\[ = - \hat{i} - 2 \hat{j} - 6 \hat{k}\]

\[\overrightarrow{BC} =\] Position vector of  C - Position vector of B
\[= 3 \hat{i} - 4 \hat{j} - 4 \hat{k} - \left( \hat{i} - 3 \hat{j} - 5 \hat{k} \right)\]
\[ = 3 \hat{i} - 4 \hat{j} - 4 \hat{k} - \hat{i} + 3 \hat{j} + 5 \hat{k} \]
\[ = 2 \hat{i} - \hat{j} + \hat{k}\]
\[\overrightarrow{CA} =\] Position vector of A- Position vector of C
\[= 2 \hat{i} - \hat{j} + \hat{k} - \left( 3 \hat{i} - 4 \hat{j} - 4 \hat{k} \right)\]
\[ = 2 \hat{i} - \hat{j} + \hat{k} - 3 \hat{i} + 4 \hat{j} + 4 \hat{k} \]
\[ = - \hat{i} + 3 \hat{j} + 5 \hat{k}\]
Clearly, 
\[\overrightarrow{AB} + \vec{BC} + \vec{CA} = \vec{0}\]
\[\text{ Now, }\overrightarrow{\left| AB \right|} = \sqrt{\left( - 1 \right)^2 + \left( - 2 \right)^2 + \left( - 6 \right)^2} = \sqrt{1 + 4 + 36} = \sqrt{41}\]
\[ \overrightarrow{\left| BC \right|} = \sqrt{\left( 2 \right)^2 + \left( - 1 \right)^2 + \left( 1 \right)^2} = \sqrt{4 + 1 + 1} = \sqrt{6}\]
\[ \overrightarrow{\left| CA \right|} = \sqrt{\left( - 1 \right)^2 + \left( 3 \right)^2 + \left( 5 \right)^2} = \sqrt{1 + 9 + 25} = \sqrt{35}\]
\[\text{ Clearly, }\overrightarrow{\left| AB \right|}^2 = \overrightarrow{\left| BC \right|}^2 + \overrightarrow{\left| CA \right|}^2 \]
\[ \Rightarrow A B^2 = B C^2 + C A^2 \]
So, A, B, C forms a right angled triangle.
shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 23: Algebra of Vectors - Exercise 23.6 [पृष्ठ ४९]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 23 Algebra of Vectors
Exercise 23.6 | Q 13 | पृष्ठ ४९

संबंधित प्रश्‍न

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


Classify the following as scalar and vector quantity.

Time period


`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 direction cosines of the vector joining the points A (1, 2, -3) and B (-1, -2, 1) directed from A to B.


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.


Write down a unit vector in XY-plane, making an angle of 30° with the positive direction of the x-axis.


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 \[\vec{a} \text{ and } \vec{b}\] \[\vec{a} = 3\hat{i} - 2\hat{j} - 6\hat{k} \text{ and } \vec{b} = 4 \hat{i} - \hat{j} + 8 \hat{k}\]


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 \[\hat{i} + \sqrt{3} \hat{j}\]


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  \[\hat{a} \text{ and } \hat{b}\] are unit vectors inclined at an angle θ, prove that \[\cos\frac{\theta}{2} = \frac{1}{2}\left| \hat{a} + \hat{b} \right|\] 


If \[\left| \vec{a} + \vec{b} \right| = 60, \left| \vec{a} - \vec{b} \right| = 40 \text{ and } \left| \vec{b} \right| = 46, \text{ find } \left| \vec{a} \right|\]


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. 


Show that the vectors \[\vec{a} = 3 \hat{i} - 2 \hat{j} + \hat{k} , \vec{b} = \hat{i} - 3 \hat{j} + 5 \hat{k} , \vec{c} = 2 \hat{i} + \hat{j} - 4 \hat{k}\] form a right-angled triangle. 


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. 


If \[\overrightarrow{AO} + \overrightarrow{OB} = \overrightarrow{BO} + \overrightarrow{OC} ,\] prove that A, B, C are collinear points.


Show that the vectors \[2 \hat{i} - 3 \hat{j} + 4 \hat{k}\text{ and }- 4 \hat{i} + 6 \hat{j} - 8 \hat{k}\] are collinear.


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.


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 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


If `veca, vecb, vecc` are vectors such that `[veca, vecb, vecc]` = 4, then `[veca xx vecb, vecb xx vecc, vecc xx veca]` =


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.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×