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**SHORT ANSWER TYPE QUESTIONS**

**Q.1. Find the unit vector in the direction of sum of vectors****Ans.**

Given that

âˆ´ Unit vector in the direction of

Hence, the required unit vector is**Q.2. Iffind the unit vector in the direction of****(i) ****(ii)****Ans.**

Given that**(i) **

âˆ´ Unit vector in the direction of

Hence, the required unit vector is**(ii)-**

âˆ´ Unit vector in the direction of

Hence, the required unit vector is**Q.3. Find a unit vector in the direction of****where P and Q have co-ordinates ****(5, 0, 8) and (3, 3, 2), respectively.****Ans.**

Given coordinates are P(5, 0, 8) and Q(3, 3, 2)

âˆ´

âˆ´ Unit vector in the direction of

Hence, the required unit vector is**Q.4. If 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.****Ans.**

Given that

BC = 1.5 BA

â‡’

â‡’

âˆ´

Hence, the required vector is**Q.5. Using vectors, find the value of k such that the points (k, â€“ 10, 3), (1, â€“1, 3) and (3, 5, 3) are collinear.****Ans.**

Let the given points are A( k , - 10 , 3), B(1,- 1, 3) and C(3, 5, 3)

If A, B and C are collinear, then

Squaring both sides, we have

âˆš

= 9 + k^{2}- 6k + 225

â‡’

Dividing by 2, we get

â‡’

â‡’

â‡’(Dividing by 2)

Squaring both sides, we get

â‡’ 10(k^{2} â€“ 2k + 82) = 784 + k^{2} â€“ 56k

â‡’ 10k^{2} â€“ 20k + 820 = 784 + k^{2} â€“ 56k

â‡’ 10k^{2} â€“ k^{2} â€“ 20k + 56k + 820 â€“ 784 = 0

â‡’ 9k^{2} + 36k + 36 = 0

â‡’ k^{2} + 4k + 4 = 0

â‡’ (k + 2)^{2} = 0

â‡’ k + 2 = 0

â‡’ k = â€“ 2

Hence, the required value is k = â€“ 2**Q.6. A vectoris inclined at equal angles to the three axes. If the magnitude of is 2 âˆš3 units, find****Ans.**

Since, the vectormakes equal angles with the axes, their direction cosines should be same

âˆ´ l = m = n

We know that

l^{2} + m^{2} + n^{2} = 1

â‡’ l^{2} + l^{2} + l^{2} = 1

â‡’

âˆ´

We know that

Hence, the required value of**Q.7. A vector has magnitude 14 and direction ratios 2, 3, â€“ 6. Find the direction cosines and components of , given that makes an acute angle with x-axis.****Ans.**

Letbe three vectors such thatand

If l, m and n are the direction cosines of vector , then

We know that l^{2} + m^{2} + n^{2 }= 1

âˆ´

âˆ´ k = Â± 2 and l =

âˆ´

â‡’

Hence, the required direction cosines areand the components of**Q.8. Find a vector of magnitude 6, which is perpendicular to both the vectors ****and****Ans.**

Letand

We know that unit vector perpendicular to

Now the vector of magnitude 6 =

Hence, the required vector is**Q.9. Find the angle between the vectors.****Ans.**

Let and let Î¸ be the angle between

âˆ´

âˆ´

Hence, the required value of Î¸ is**Q.10. If show that Interpret the result geometrically?****Ans.**

Given that

So,

â‡’

â‡’

â‡’

â‡’...(i)

Now

â‡’

â‡’

â‡’

â‡’

âˆ´...(ii)

From eq. (i) and (ii) we get

Hence proved.

Geometrical Interpretation

According to figure, we have

Area of parallelogram ABCD is

Since, the parallelograms on the same base and between the same parallel lines are equal in area**Q.11. Find the sine of the angle between the vectors ****and****Ans.**

Given that

We know that

âˆ´

âˆ´

â‡’

Hence,**Q.12. If A, B, C, D are the points with position vectors****respectively, find the projection of****Ans.**

Here, Position vector of A =

Position vector of B =

Position vector of C =

Position vector of D =

Projection of

Hence, the required projection = âˆš21 .**Q.13. Using vectors, find the area of the triangle ABC with vertices A(1, 2, 3), B(2, â€“ 1, 4) and C(4, 5, â€“ 1).****Ans.**

Given that A(1, 2, 3), B(2, â€“1, 4) and C(4, 5, â€“1)

Area of Î”ABC ==

=

Hence, the required area is**Q.14. Using vectors, prove that the parallelogram on the same base and between the same parallels are equal in area.****Ans.**

Let ABCD and ABFE be two parallelograms on the same base AB and between same parallel lines AB and DF.

Let

âˆ´ Area of parallelogram ABCD =

Now Area of parallelogram ABFE =

Hence proved.

**LONG ANSWER TYPE QUESTIONS**

**Q.15. Prove that in any triangle ABC,where a, b, c are the magnitudes of the sides opposite to the vertices A, B, C, respectively.****Ans.**

Here, in the given figure, the components of c are c cos A and c sin A.

âˆ´

In DBDC,

a^{2} = CD^{2} + BD^{2}

â‡’ a^{2} = (b â€“ c cos A)^{2 }+ (c sin A)^{2}

â‡’ a^{2} = b^{2} + c^{2} cos^{2} A â€“ 2bc cos A + c^{2} sin^{2} A

â‡’ a^{2} = b^{2} + c^{2}(cos^{2} A + sin^{2} A) â€“ 2bc cos A

â‡’ a^{2} = b^{2} + c^{2} â€“ 2bc cos A

â‡’ 2bc cos A = b^{2} + c^{2} - a^{2}

âˆ´

Hence Proved.**Q.16. Ifdetermine the vertices of a triangle, show that****gives the vector area of the triangle. Hence deduce the ****condition that the three points** **are collinear. Also find the unit vector normal to the plane of the triangle.****Ans.**

Since,are the vertices of Î”ABC

For three vectors are collinear, area of Î”ABC = 0

which is the condition of collinearity of

Letbe th e unit vector normal to the plane of the Î”ABC

âˆ´

â‡’**Q.17. Show that area of the parallelogram whose diagonals are given by ****is****. Also find the area of the parallelogram whose diagonals are ****and****Ans.**

Let ABCD be a parallelogram such that,

âˆ´ by law of triangle, we get

Adding eq. (i) and (ii) we get,

Subtracting eq. (ii) from eq. (i) we get

So, the area of the parallelogram ABCD =

Now area of parallelogram whose diagonals areand

Hence, the required area is**Q.18. Iffind a vector such that ****Ans.**

Let

Also given that

Since,

âˆ´

On comparing the like terms, we get

c_{3} â€“ c_{2} = 0 ...(i)

c_{1} â€“ c_{3} = 1 ...(ii)

and c_{2} â€“ c_{1} = â€“1 ...(iii)

Now

âˆ´ c_{1} + c_{2} + c_{3} = 3 ...(iv)

Adding eq. (ii) and eq. (iii) we get,

c_{2} â€“ c_{3} = 0 ...(v)

From (iv) and (v) we get

c_{1} + 2c_{2} = 3 ...(vi)

From (iii) and (vi) we get

Adding

âˆ´

c_{3} â€“ c_{2} = 0

â‡’

âˆ´

Now c_{2} â€“ c_{1} = â€“ 1 â‡’

â‡’

âˆ´

Hence,

**OBJECTIVE TYPE QUESTIONS**

**Q.19. The vector in the direction of the vectorthat has magnitude 9 is****(a)****(b)****(c)****(d)****Ans. **(c)**Solution.**

Let

Unit vector in the direction of

âˆ´ Vector of magnitude 9 =

Hence, the correct option is (c).**Q.20. The position vector of the point which divides the join of points****in the ratio 3 : 1 is****(a)****(b)****(c)****(d)****Ans. **(d)**Solution.**

The given vectors areand the ratio is 3 : 1.

âˆ´ The position vector of the required point c which divides the join of the given vectors

Hence, the correct option is (d).**Q.21. The vector having initial and terminal points as (2, 5, 0) and (â€“3, 7, 4), respectively is****(a)****(b)****(c)****(d)****Ans. **(c)**Solution.**

Let A and B be two points whose coordinates are given as (2, 5, 0) and (â€“ 3, 7, 4)

âˆ´

â‡’

Hence, the correct option is (c).**Q.22. The angle between two vectors****with magnitudes âˆš3 and 4, respectively, ****and****(a) Ï€/6****(b) Ï€/3****(c) Ï€/2****(d) 5Ï€/2****Ans. **(b)**Solution.**

Here, given that

âˆ´ From scalar product, we know that

â‡’

â‡’

âˆ´

Hence, the correct option is (b).

Q.23. Find the value of Î» such that the vectors**are orthogonal****(a) 0 ****(b) 1 ****(c) 3/2****(d) -5/2****Ans. **(d)**Solution.**

Given that

Sinceare orthogonal

âˆ´

â‡’ 2 + 2Î» + 3 = 0

â‡’ 5 + 2Î» = 0 â‡’

Hence, the correct option is (d).**Q.24. The value of Î» for which the vectors****are parallel is****(a) 2/3****(b) 2/3****(c) 5/2****(d) 2/5****Ans. **(a)

Solution.

Let

Since the given vectors are parallel,

âˆ´ Angle between them is 0Â°

so

Squaring both sides, we get

900 + Î»^{2} + 60Î» = 46(20 + Î»^{2})

â‡’ 900 + Î»^{2} + 60Î» = 920 + 46Î»^{2}

â‡’ Î»^{2} â€“ 46Î»^{2} + 60Î» + 900 â€“ 920 = 0

â‡’ - 45Î»^{2} + 60Î» - 20 = 0

â‡’ 9Î»^{2} â€“ 12Î» + 4 = 0

â‡’(3Î» â€“ 2)^{2} = 0

â‡’3Î» â€“ 2 = 0

â‡’ 3Î» = 2

âˆ´ Î» = 2/3**Alternate method:**Let

If

âˆ´

â‡’

Hence, the correct option is (a).

Solution.

Let O be the origin

âˆ´

and

âˆ´ Area of Î”OAB =

Hence the correct option is (d).

Solution.

Let

âˆ´

Now,

âˆ´

Similarly

and

Hence, the correct option is (d).

(a) 5

Solution.

Given that

âˆ´

â‡’ 12 = 10 Ã— 2 Ã— cos Î¸

â‡’

âˆ´

â‡’

â‡’

Now

Hence, the correct option is (d).**Q.28. The vectorsare coplanar if****(a) Î» = â€“2 ****(b) Î» = 0 ****(c) Î» = 1 ****(d) Î» = â€“ 1****Ans. **(a)

Solution.

Let

Ifare coplanar, then

âˆ´

â‡’ l(Î»^{2} â€“ 1) â€“ 1 (Î» + 2) + 2(â€“1 â€“ 2l) = 0

â‡’ Î»^{3} â€“ Î» â€“ Î» â€“ 2 â€“ 2 â€“ 4Î» = 0

â‡’ Î»^{3} â€“ 6Î» â€“ 4 = 0

â‡’ (Î» + 2) (Î»^{2} â€“ 2Î» â€“ 2) = 0

â‡’ Î» = â€“ 2 or Î»^{2} â€“ 2Î» â€“ 2 = 0

â‡’

â‡’

âˆ´

Hence, the correct option is (a).**Q.29. Ifare unit vectors such that****then the value of**

**(a) 1 (b) 3 (c) -3/2**

Solution.

Given that

and

âˆ´

â‡’

â‡’

â‡’

â‡’

Hence, the correct option is (c).

Solution.

The projection vector of

Hence, the correct option is (a).

Solution.

Given that

and

â‡’

â‡’

â‡’

â‡’

â‡’

â‡’

Hence, the correct option is (c).

(b) [â€“ 12, 8]

(c) [0, 12]

(d) [8, 12]

Solution.

Given that

Now

Here - 3 â‰¤ Î» â‰¤ 2

â‡’ - 3.4 â‰¤ 4Î» â‰¤ 2.4

â‡’ - 12 â‰¤ 4Î» â‰¤ 8

âˆ´ 4Î» = [- 12, 8]

Hence, the correct option is (b).

Solution.

The number of vectors of unit length perpendicular to vectors

âˆ´

So, there will be two vectors of unit length perpendicular to vectors

Hence, the correct option is (b).

**FILL IN THE BLANKS**

**Q.34. The vectorbisects the angle between the non-collinear vectors****if ________.****Ans.**

If vector bisects the angle between non-collinear vectors

then the angle betweenis equal to the angle between

So, ...(i)

Also,[âˆµ Î¸ is same] ...(ii)

From eq. (i) and eq. (ii) we get,

â‡’

â‡’

Hence, the required filler is**Q.35. Iffor some non-zero vector****then the value of ****is ________****Ans.**

Ifis a non-zero vector, thencan be in the same plane.

Since angles between and are zero i.e. Î¸ = 0

Hence the required value is 0.**Q.36. The vectorsa re the adjacent sides of a parallelogram. The acute angle between its diagonals is ________.****Ans.**

Given that

and

Let Î¸ be the angle between the two diagonal vectors

then

Hence the value of required filler is**Q.37. The values of k for which ****is parallel to****holds true are _______.****Ans.**

Given that

Now sinceis parallel to

Here we see that atbecome null vector and then it will not be

parallel to

Now sinceis parallel to

Here we see that atbecome null vector and then it will not be

parallel to

Hence, the required value of k âˆˆ (- 1, 1) and k â‰ **Q.38. The value of the expressionis _______.****Ans.**

Hence, the value of the filler is**Q.39. Ifand****is equal to _______.****Ans.**

â‡’

â‡’

â‡’

â‡’

â‡’

â‡’

âˆ´

Hence, the value of the filler is 3.**Q.40. Ifany non-zero vector, then****equals _______.****Ans.**

Let

âˆ´

= a_{1}

Hence, the value of the filler is**State True or False in each of the following Exercises.****Q.41. Ifthen necessarily it implies****Ans.**

Ifthenwhich is true.

Hence, the statement is True.

Q.42. Position vector of a point P is a vector whose initial point is origin.**Ans.**

True

Q.43. Ifthen the vectors**are orthogonal.****Ans.**

Given that

Squaring both sides, we get

â‡’

â‡’

â‡’

which implies thatare orthogonal.

Hence the given statement is True.**Q.44. The formula****is valid for non-zero vectors****Ans.**

Hence, the given statement is False.**Q.45. Ifare adjacent sides of a rhombus, then****Ans.**

If

So the angle between the adjacent sides of the rhombus should be 90Â° which is not possible.

Hence, the given statement is False.

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