EasyQ1. If and the resultant of & is (4, 5), find (5,1)(1,5)(-1,-5)(-1,5)(-5,1)Solution = - +4 ……………. (1) += - +4 ……….. (2)Putting the value of from equation (1) into (2) we get +4 + =4 +5 4 + +5 - 4 = 5 + Thus is (5, 1)Q2.if is (6,-4). Find its norm?2√1313√212√22√12None of theseSolution For any vector = a +b the norm is defined as || = Thus || = = = √52 = 2√13Q3. If = 2 +3 & = - 5 the resultant vector 3 +2 is equal to 7 - 9 5 + 3 - 2 8 - 2 +3 Solution 3 = 3( -5 ) = 3 – 15 2 =2( 2 +3 ) = 4 + 6 3 += 3 -15 +4 + 6 = 7 - 9 Q4. Let = (2, 3) & = (6,-4) the find the dot product of & ?-1012-2Solution =6 – 4 = 2 +3 * =(2 +3 ) * ( 6 – 4 ) = 12 – 12 = 0As * = * = * = 1 * = * = * = 0Q5. Find unit vector in the direction of vector 2 + + + + + + + + - + - - Solution The unit vector in the direction of a vector is given by = | = = √14Therefore unit vector is = + - ) = + + Q6. The magnitude of vector + + is √33√22None of theseSolutionLet the vector x + +zThen its magnitude is given by | = Thus for + + | = = √3Q7. Find the position vector of the midpoint of vector joining the point p (2,3,4) & q (4,1,-2).(3,2,1)(1,2,3)(2,1,3)(3,1,2)(1,1,2)Solution The midpoint for two points (x1 , y1 ,z1) & (x2 , y2 ,z2) is given by() Thus the midpoint of vector joining p (2,3,4) & q (4,1,-2) is ( , , ) = (3,2,1)Q8.find the angle between two vector & with magnitude 1 & 2 respectively & when * = 1 ΠSolution Given * = 1 | = 1 & | |= 2 * =| | | = = Ѳ = = 600 = Q9. Find - | , if two vector & are such that | = 2 & | | = 3 & √5√412Cannot be determinedsolutionWe have - |2 =( - )* ( - ) = * - * - * + * = |2 – 2 ( * ) +|2 = (2)2 – 2 *(4) + (3)2 = 5- | = √5Q10. If - -3 & +3 -5 then + is equal to -2 +8 +2 - 8 - -8 - - 4 + +4 Solution + = - -3 ) + +3 -5 ) = +2 - 8 4 3 2 1-1 1 2 3-4-3-2-1MediumQ1. If - -3 & +3 -5 then - is equal to -4 +2 -4 +2 +4 +2 +2 +2 - + Solution- = - -3 - +3 -5 ) = - -3 -3 +5 = -4 +2 Q2.vector has component (8,15) (3,3).if - what is the magnitude of vector?10.513.015.616.521.1Solution = 8 + 15 = 3 + 3 - = 8 + 153 - 3 = 5 + 12Thus the magnitude ofis = = = = 13ѲQuestion Fig400Answer FigQ3. In the figure above, the magnitude of vector 9 and the magnitude of vector is is 7 ,what is the magnitude of vector () ?5.7911.407.0012.2615.05SolutionLet and be two vectors having angle Ѳ between the resultant = = 9 = 7 and cos Ѳ 400= Q4. Find the angle between the vectors - 2+3 and - 2 + None of theseSolutionLet = - 2 +3 = 3 - 2 + . cosѲ = = = √14 = = = √14Cos Ѳ = = = = 10/14 = 5/7 Ѳ=Q5. If = + 2+3 and λ =-2 find the value of λ= + 4+4 -4 - 4 +8+ 6 - 2-2 - +3/2 Solutionλ=(-2)( + 2+3 ) = -4 - 4Q6. if is unit vector & () * (+) = 8 then find ||98123Cannot be determinedSolution Since is unit vector . | = 1 () * (+) = 8 * + * - * - * = 8||2 - ||2 = 8 ||2 – 1 = 8 ||2 = 9|| = 3 (as magnitude of vector in non negative)Q7 if = + then what is the magnitude of ?3√55√3√56√652Solution As has co-ordinate (-3,4) & b has co-ordinate (-3,-1) -4 - = + = -4 - = +3 || = = = √45 = 3√5Q8. If +2 +3 & +2 +are perpendicular to each other?3.5-3.55.3-5.3Cannot be determinedSolutionWhen & are perpendicular to each other their product is zero . 0 ( +2 +3 ) * ( +2 + ) = 0 2x +4+3 = 02x +7 = 02x = -7X = -3.5Q9. If & are unit vector then ( -5 ) * ( + ) is -29 +11 * -29 -11 * Cannot be determined-11 * 11 * Solution As & are unit vector| | = | | = 1( -5 ) * ( + )=6|2 + 2 ( * ) -10 & -35 |2 = 6 -35 + 11 * =-29 +11 * Q10. If + + & -3 - then the value of-2 - + + -9 -5 + + +2 + - + Solution = 3( + + ) = +3 +3 2 = 2( -3 -) = -6 --2 = +3 +3 - ( -6 -) = +3 +3 - +6 + = - + + Hard Q1. If +2 +3 & +2 + & + are such that +λ is perpendicular to, then find the value of λ?1.52.54.56.27.3Solution +λ = +2 +3 +λ ( +2 +) = + (2+2λ) + (3+λ) + As +λ is perpendicular to, so the dot product is zero +λ * = 0 [ + (2+2λ) + (3+λ) ]* + ) = 0 + (3+λ) = 0 6-3λ + 3+ λ = 0 9- 2λ = 0 2 λ = 9 .λ = 4.5Q2. Find the magnitude of two vectors & having same magnitude & such that the angle between them is 600 & their scalar product is ½.1-1021/Solution * = (given) | | = | | (given) = | | | | Ѳ = 600| |2 = = | |2 = = 1 | | = 1Q3. The projection of the vector +3 +2 on the vector +2 + is Cannot be determinedSolution The projection of vector on the vector is given by ( * ) | | = = √6* = +3 +2 ) * +2 +) = 2+6+2 = 10 The projection of vector on Vector = Q4. Find the value of x & y so that the vector +3 & +y are equalx=1 & y=3x=3& y=2x=-2&y=3x=2& y=-3data insufficientSolution +3 = +y X = 2 & y =3Q5.the direction ratio’s of the vector + -2 is (, , )(1,1,-2)(, , )(4,-3, 2)(1,1,-1)Solution The direction ratio’s a , b & c of a vector +y + are the respective of x , y & z of vector , so for a given vector , we have a=1 , b=1 & c = -2.If l, m & n are the direction cosine of the given vector then .l = = m = = n = = As | | = = √6Thus the cosine are (, , ) Q6. The unit vector in the direction of the sum of the vectors +2 -5 & + + ( +3 -2) ( +3 -2) ( -3 -2) ( +4 +2) ( +3 +4)Solution The sum of the given vector is = + = +3 -2| = = √29Thus unit vector is = = ( +3 -2)Q7.if + + & - - then the dot product of & is 01 + None of theseSolution + + & - - * =( + + )* - -) = 2-1-1 = 2-2 = 0 [ * = * = * ] [ * = * = * ]Q8. Find the norm of + + is √14712212SolutionIf the vector is +b +c The norm given by Here a=2 b=1 c=3Thus norm of vector is = = = Q9. If the magnitude of &| are equal & angle between them is 600. The resultant of the two vectors is 1. Find the magnitude of vectors31Cannot be determinedSolution the magnitude of &| are equal thus =| = xR = Where R is resultant1 = = 1 = 1 = X2= X = Q10. The unit vector has magnitude of0125No such vector existSolutionA unit vector is a vector whose magnitude is 1.