English

If → a × → B = → C × → D and → a × → C = → B × → D , Show that → a − → D is Parallel to → B − → C Where → a ≠ → D and → B ≠ → C . - Mathematics

Advertisements
Advertisements

Question

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 \[\vec{a} \neq \vec{d} \text { and } \vec{b} \neq \vec{c}\] .

Advertisements

Solution

Two non zero vectors are parallel if and only if their cross product is zero vector.
So, we have to prove that cross product of \[\vec{a} - \vec{d}\] and  \[\vec{b}  -  \vec{c}\] is zero vector. \[\left( \vec{a} - \vec{d} \right) \times \left( \vec{b} - \vec{c} \right) = \left( \vec{a} \times \vec{b} \right) - \left( \vec{a} \times \vec{c} \right) - \left( \vec{d} \times \vec{b} \right) + \left( \vec{d} \times \vec{c} \right)\]

Since, it is given that 

\[\vec{a} \times \vec{b} = \vec{c} \times \vec{d} \text { and } \vec{a} \times \vec{c} = \vec{b} \times \vec{d}\]

And,

\[\vec{d} \times \vec{b} = - \vec{b} \times \vec{d} \]

\[ \vec{d} \times \vec{c} = - \vec{c} \times \vec{d}\]

Therefore, 

\[\left( \vec{a} - \vec{d} \right) \times \left( \vec{b} - \vec{c} \right) = \left( \vec{a} \times \vec{b} \right) - \left( \vec{a} \times \vec{c} \right) - \left( \vec{d} \times \vec{b} \right) + \left( \vec{d} \times \vec{c} \right)\]

\[ = \left( \vec{c} \times \vec{d} \right) - \left( \vec{b} \times \vec{d} \right) + \left( \vec{b} \times \vec{d} \right) - \left( \vec{c} \times \vec{d} \right)\]

\[ = \vec{0}\]

Hence,

\[\vec{a} - \vec{d}\] is parallel to 

\[\vec{b} - \vec{c}\] where  \[\vec{a}  \neq  \vec{d}   \text { and }     \vec{b}  \neq  \vec{c}\] .

shaalaa.com
  Is there an error in this question or solution?
2015-2016 (March) Foreign Set 2

RELATED QUESTIONS

Find the direction ratios of a vector perpendicular to the two lines whose direction ratios are -2, 1, -1, and -3, -4, 1.


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.


Classify the following measures as scalar and vector.

10 kg


Classify the following as scalar and vector quantity.

Time period


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.


Express \[\vec{AB}\]  in terms of unit vectors \[\hat{i}\] and \[\hat{j}\], when the points are A (−6, 3), B (−2, −5)
Find \[\left| \vec{A} B \right|\] in each case.


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 angles which the vector \[\vec{a} = \hat{i} -\hat {j} + \sqrt{2} \hat{k}\] makes with the coordinate axes.


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


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 \[\vec{p} = 5 \hat{i} + \lambda \hat{j} - 3 \hat{k} \text{ and } \vec{q} = \hat{i} + 3 \hat{j} - 5 \hat{k} ,\] then find the value of λ, so that \[\vec{p} + \vec{q}\] and \[\vec{p} - \vec{q}\]  are perpendicular vectors. 


If \[\vec{\alpha} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k} \text{ and } \vec{\beta} = 2 \hat{i} + \hat{j} - 4 \hat{k} ,\] 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 \[\vec{a} = 2 \hat{i} + 2 \hat{j} + 3 \hat{k} , \vec{b} = - \hat{i} + 2 \hat{j} + \hat{k} \text{ and } \vec{c} = 3 \hat{i} + \hat{j}\] \[\vec{a} + \lambda \vec{b}\] is perpendicular to \[\vec{c}\] then find the value of λ. 


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 unit vector in the direction of vector \[\overrightarrow{PQ} ,\]

 where P and Q are the points (1, 2, 3) and (4, 5, 6).


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 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


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.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×