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501.

If vectors A and B be respectively equal to `3hati - 4hatj + 5hatk and 2hati + 3hatj - 4hatk.` Find the unit vector parallel to A + B

Answer» `hatn=(vecA+vecB)/(|vecA+vecB|)=((3hati-4hatj+5hatk)+(2hati+3hatj-4hatk))/(5hati-hatj+hatk)`
`=((5hati-hatj+hatk))/sqrt(27)`
Discussion
502.

Deduce the condition for the vectors `2hati + 3hatj - 4hatk and 3hati - alpha hatj + bhatk` to be parallel.A. `a=-9//2,b=-6`B. `a=-6,b=-9//2`C. `a=4,b=5`D. `a=8,b=2`

Answer» Correct Answer - A
`(x_(1))/(x_(2))=(y_(1))/(y_(2))=(z_(1))/(z_(2))`
Discussion
503.

Let `vec(C )= vec(A)+vec(B)` thenA. `|vec(C )|` is always greater than `|vec(A)|`B. It is possible to have `|vec(C )|` lt |vec(A)|` and |vec(C )| lt |vec(B)|`C. `C` is always equal to `A + B`D. `C` is never equal to `A + B`

Answer» Correct Answer - B
Discussion
504.

A particle is being acted upon by four forces of `30 N` due east , `20 N` due north , `50 N` due west and `40 N` due south. The resultant force will beA. `20 sqrt(2) N , 60^(@)` south to westB. `20 sqrt(2) N , 45^(@)` south westC. `20 sqrt(2) N , 45^(@)` north to eastD. `20 sqrt(2) N , 45^(@)` south to east

Answer» Correct Answer - B
Discussion
505.

A particle is being acted upon by four forces of `30 N` due east , `20 N` due north , `50 N` due west and `40 N` due south. The resultant force will be

Answer» Correct Answer - `20 sqrt(2)N, 45^(@)` south of west
Discussion
506.

If `|a|=5, |b|6 and a.b=-25`, then `|axxb|` is equal toA. 25B. `6sqrt(11)`C. `11sqrt5`D. `5sqrt(11)`

Answer» Correct Answer - D
Now ,` |axxb|^2+(a.b)^2=|a|^2|b|^2`
`rArr |axxb|^2=25xx36-(25)^2`
`=25(36-25)=25xx11`
`rArr |axxb|=5sqrt(11)`
Discussion
507.

Which one of the following vectors is a magnitude 6 and perpendicular to both `a=2hati+2hatj+hatk and b=hati-2hatj+2hatk`?A. `2hati-hatj-2hatk`B. `2(2hati-hatj+2hatk)`C. `3(2hati-hatj-2hatk)`D. `2(2hati-hatj-2hatk)`

Answer» Correct Answer - D
Now , `axxb=|{:(hat(i),hat(j),hat(k)),(2,2,1),(1,-2,2):}|=hat(i)(4+2)-hat(j)(4-1)+hat(k)(-4-2)`
`6hat(i)-3hat(j)-6hat(k)`
`|axxb|=sqrt(36+9+36)=sqrt(81)=9`
`therefore` Requried vectors are `+-6|(axxb)/(|axxb||)|=+-(6)/(9)(6hat(i)-3hat(j)-6hat(k))=+-2(2hat(i)-hat(j)-2hat(k))`
Discussion
508.

Find a vector ` vec r`of magnitude `3sqrt(2)`units which makes an angle of `pi/4`and `pi/2`with `y`and z-axis respectively.A. `r= pm 3hati+3hatj`B. `r=3hati+3hatj`C. `r=pm 3hati+3hatj`D. None of these

Answer» Correct Answer - A
Here, `m="cos"pi/4=1/sqrt(2)` and `n="cos"pi/2 =0`
Therefore, `l^(2)+m^(2)+n^(2)=1` [given]
`l^(2)+1/2 +0=1`
`implies l= pm 1/sqrt(2)`
Hence, the required vector `r=3sqrt(2) (l hat(i)+m hat(j)+n hat(k))` is given by `r=3sqrt(2) (pm 1/sqrt(2) hat(i) +1/sqrt(2) hat(j) + 0hat(k))=r= pm 3 hat(i)+3hat(j)`
Discussion
509.

If vector `P, Q and R` have magnitude 5,12,and 13 units and `vec(P)+vec(Q)=vec(R )`, the angle between Q and R isA. `"cos"^(-1)5/12`B. `"cos"^(-1)5/13`C. `"cos"^(-1)12/13`D. `"cos"^(-1)7/13`

Answer» Correct Answer - C
Discussion
510.

If P + Q = R and P - Q = S, prove that `R^2 + S^2 = 2(P^2 + Q^2)`

Answer» `R^2 = P^2 Q^2 + 2PQ cos theta …..(i)`
`S^2 = P^2 + Q^2 - 2PQ cos theta (i)`
Adding these two equations we get,
`R^2 + S^2 = 2(P^2 +Q^2)`
Discussion
511.

The resultant of two vector A and B is at right angles to A and its magnitude is half of B. Find the angle between A and B.A. `120^(@)`B. `150^(@)`C. `135^(@)`D. None of these

Answer» Correct Answer - B
`B/2= sqrt(A^(2)+B^(2)+2AB cos theta)`…..(i)
`:. Tan90^(@)=(B sin theta)/(A+B cos theta)implies A+B cos theta=0`
`:. cos theta= -(A)/(B)`
Hence, from (i), `(B^(2))/4= A^(2)+B^(2)-2A^(2)implies A= sqrt(3)B/2`
`implies cos theta = -(A)/(B)= -(sqrt(3))/2, :. theta= 150^(@)`
Discussion
512.

If two vectors are `A = 2hati + hatj - hatk and B = hatj - 4hatk.` By calculation, prove AxxB isperpendicular to both A and B.

Answer» Find `A xx B` and then prove that` (A xx B) = 0` and
`(A xx B) .B =0`.
it means `(AxxB) _|_` to both `A` and `B` .
Discussion
513.

Let ` vec u , vec va n d vec w`be such that `| vec u|=1,| vec v|=2a n d| vec w|=3.`If the projection of ` vec v`along ` vec u`is equal to that of ` vec w`along ` vec u`and vectors ` vec va n d vec w`are perpendicular to eachother, then `| vec u- vec v+ vec w|`equals`2`b. `sqrt(7)`c. `sqrt(14)`d. `14`A. 2B. `sqrt7`C. `sqrt(14)`D. 14

Answer» Correct Answer - C
Since, `|u| = 1, |v| = 2,|w| = 3`
The projection of v along `u = (v.u)/(|u|)` and the projection of w along `u = (w.u)/(|u|)`.
According to given condition.
`(v.u)/(|u|) = (w.u)/(|u|)`
`rArr v.u = w.u "…."(i)`
Since, v and w are perpendicular to each other.
`:. v. w = 0`
Now, `| u- v + w|^(2) = |u|^(2) + |v|^(2) +|w|^(2)`
`- 2u . v - 2v.w+2u.w`
`rArr |u- b- v + w|^(2) = 1 + 4+ 9 - 2 u . v + 0 + 2u . v` [from Eq. (i) ]
`rArr |u - v + w|^(2) = 1 + 4 + 9`
`:. |u- v + w| = sqrt(14)`
Discussion
514.

If the projection of the vector a on b is `|axxb|` and if `3b=hati+hatj+hatk`, then the angle between a and b isA. `(pi)/(3)`B. `(pi)/(2)`C. `(pi)/(4)`D. `(pi)/(6)`

Answer» Correct Answer - A
Given ,projection of `a on b=|axxb|`
`rArr (a.b)/(|b|)=|axxb|`
`rArr (|a||b|costheta)/(|b|)=|a||b|sintheta`
`rArr tan theta=(1)/(|b|)`
`rArr tan theta= (1)/((1)/(3)sqrt(1^2+1^2+1^2))`
`rArr tan theta = sqrt(3)`
`rArr theta=(pi)/(3)` .
Discussion
515.

`(a.(b xx c))/(b.(c xx a))+(b.(a x b))/(a. (b xx c))` is equal toA. 1B. 2C. 0D. `oo`

Answer» Correct Answer - B
Now, `(a*(bxxc))/(b*(cxxa))+(b*(axxb))/(a*(axxc))=([abc])/(["bca"])+([bab])/(["abc"])`
`=1+0=1" "[because[bab]=0]`
Discussion
516.

Let ` vec a=2 hat i- hat j+ hat k , vec b= hat i+2 hat j= hat ka n d vec c= hat i+ hat j-2 hat k`be three vectors. A vector in the plane of ` vec ba n d vec c`, whose projection on ` vec a`is of magnitude `sqrt(2//3)`, is`2 hat i+3 hat j-3 hat k`b. `2 hat i-3 hat j+3 hat k`c. `-2 hat i- hat j+5 hat k`d. `2 hat i+ hat j+5 hat k`A. `2hat(i)+3hat(j)-3hat(k)`B. `2hat(i)+3hat(j)+3hat(k)`C. `2hat(i)-5hat(j)+5hat(k)`D. `2hat(i)+hat(j)+5hat(k)`

Answer» Correct Answer - A
Any vector I the place of b and c is
`r=mb+c(m+1)hati+(2m+1)hati+(-m-2)hatk`
Projection of r on a `=(r*a)/(|a|)=|sqrt(2/3)|`
`:." "(2(m+1)-(2m+1)+(-m-2))/(sqrt6)=+-sqrt(2/3)`
`rArr" "-m-2+-2rArrm=-3"and"1`
Hence, `r=-2hati-5hatj+hatk" and "r=2hati+3hatj-3hatk`
Discussion
517.

Vectors ` vec a=-4 hat i+3 hat k ; vec b=14 hat i+2 hat j-5 hat k`are laid off fromone point. Vector ` hat d`, which is being laid offrom the same point dividing the angle between vectors ` vec aa n d vec b`in equal halves and having the magnitude `sqrt(6),`isa. ` hat i+ hat j+2 hat k`b.` hat i- hat j+2 hat k`c. ` hat i+ hat j-2 hat k`d. `2 hat i- hat j-2 hat k`A. `hati+hatj + 2hatk`B. `hati-hatj+2hatk`C. `hati+hatj-2hatk`D. `2hati-hatj-2hatk`

Answer» Correct Answer - A
`hat a = (-4i + 3hatk)/(5) , hatb = (14hati + 2hatj - 5hatk)/(15) `
A vector `vecV` bisecting the angle between `veca and vecb` is
`" "vecV = hata + hat b `
`" "= (-12hati + 9hatk + 14hati + 2hatj -5hatk)/(15)`
`" " = (2hati + 2hatj + 4hatk)/(15)`
Required vector `vecd = sqrt6 hat V = hat i + hatj + 2hatk`
Discussion
518.

If a, b and c are three non-zero vectors such that each one of them being perpendicular to the sum of the other two vectors, then the value of `|a+b+c|^(2)` isA. `|a|^(2)+|b|^(2)+|c|^(2)`B. `|a|+|b|+|c|`C. `2 (|a|^(2)+|b|^(2)+|c|^(2))`D. `1/2 (|a|^(2)+|b|^(2)+|c|^(2))`

Answer» Correct Answer - A
According to the given condition, each vectors is perpendicular to the sum of the vectors.
`:. a.(b+c) =0`,
`b.(a+c) =0`
and `c.(a+b) = 0`
`rArr a.b+a.c = 0, b.a + b.c = 0`
and `c.a + c.b = 0`
`rArr 2(a.b+b.c +c.a) = 0`
`rArr a.b+b.c+c.a= 0"...."(i)`
Now, `|a+b+c|^(2) = |a|^(2) + |b|^(2) + |c|^(2) + 2(a.b+b.c+c.a)`
`= |a|^(2) +|b|^(2) +|c|^(2) + 2(0)` [from eq. (i)]
`= |a|^(2) +|b|^(2) +|c|^(2)`
Discussion
519.

One of the rectangular components of velocity of `80Kmh^(-1) is 40Kmh^(-1)`. What is the other components?A. `40Kmh^(-1)`B. `69.28Kmh^(-1)`C. `89.44Kmh^(-1)`D. `120Kmh^(-1)`

Answer» Correct Answer - B
`R= 80 km//h, R_(x)= 40 km//h`
but `R^(2)= R_(x)^(2)+R_(y)^(2)implies 80^(2)= 40^(2)+R_(y)^(2)`
`implies R_(y)^(2)= 4800`
`implies R_(y)= sqrt(4800)= 40 sqrt(3)= 69.28 km//h`
Discussion
520.

Find the resultant of vectors `veca=hati-hatj+2hatk and vecb=hati+2hatj-4hatk`. Find the unit vector in the direction of the resultant vector.

Answer» Let the resultant vector of `veca and vecb` is `veca+vecb=2hati+hatj-2hatk=vecc`
Now unit vector in the direction of `vecc` is `hatc=(2hati+hatj-2hatk)/(sqrt((2)^(2)+(1)^(2)+(-2)^(2))=(1)/(3)(2hati+hatj-2hatk)`
Discussion
521.

Find the angle of vector `veca=6hati+2hatj-3hatk` with `x`-axis.

Answer» `" "veca=6hati+2hatj-3hatk`
or `" "|veca|=sqrt((6)^(2)+(2)^(2)+(-3)^(2))=7`
Hence, the angle of vector with the `x`-axis is `cos^(-1)""(6)/(7)`.
Discussion
522.

Check whether the three vectors `2hati+2hatj+3hatk, vecb=-3hati+3hatj+2hatk and vecc=3hati+4hatk` form a triangle or not.

Answer» If vectors `veca=2hati+2hatj+3hatk, vecb=-3hati+3hatj+2hatk` and `vecc=3hati+4hatk` form a triangle, then we must have
`" "veca+vecb+vecc=0`
But for given vectors, `veca+vecb+vecc ne 0`. Hence, these vectors do not form a triangle.
Discussion
523.

If `vec2hati+3hatj` and `vecB=2hatj+3hatk` the component of `vecB` along `vecA` isA. 6B. `(1)/(6)`C. `(6)/(13)`D. `(6)/sqrt(13)`

Answer» Correct Answer - D
`b cos theta=(veca.vecb)/(a)`
Discussion
524.

if the vectors `P = alphahati+alphaj+3hatk` and `Q = alphahati - 2 hatj - hatk` are perpendicular to each other, then the positive value of `alpha` isA. ZeroB. 1C. 2D. 3

Answer» Correct Answer - D
`vecA.vecB=a^(2)-2a-3=0rArr(a-3)(a+1)=0`
Discussion
525.

If `|vec(A)xxvec(B)|=sqrt(3)vec(A).vec(B)`, then the value of `|vec(A)+vec(B)|` isA. `(A^(2)+B^(2)+(AB)/(sqrt(3)))^(1//2)`B. `A+B`C. `(A^(2)+B^(2)+sqrt(3)AB)^(1//2)`D. `(A^(2)+B^(2)+AB)^(1//2)`

Answer» Correct Answer - D
`|vec(A)xxvec(B)|=sqrt(3)(vec(A).vec(B))`
`AB sin theta= sqrt(3) AB cos theta implies tan theta =sqrt(3):. theta = 60^(@)`
Now, `|vec(R )|=|vec(A)|+vec(B)|= sqrt(A^(2)+B^(2)+2AB cos theta)`
`=sqrt(A^(2)+B^(2)+2AB(1/2))= (A^(2)+B^(2)+AB)^(1//2)`
Discussion
526.

If `|vec(A)xxvec(B)|=sqrt(3)vec(A).vec(B)`, then the value of `|vec(A)+vec(B)|` isA. `(A^(2) + B^(2) + (AB)/(sqrt(3)))^(1//2)`B. `A + B`C. `(A^(2) + B^(2) + sqrt(3) AB)^(1//2)`D. `(A^(2) + B^(2) + AB)^(1//2)`

Answer» Correct Answer - D
`|vec(A) xx vec(B)| = sqrt(3) vec(A) . Vec(B)`
`AB sin theta = sqrt(3) AB cos theta`
`tan theta = sqrt(3) rArr theta = 60^(@)`
`|vec(A) + vec(B)| = sqrt(A^(2) + B^(2) + 2 AB cos 60^(@))`
` = (A^(2) + B^(2) + AB )^(1//2)`
Discussion
527.

The scalar product of two vectors is `2 sqrt(3)` and the magnitude of their vector product is `2`. The angle between them isA. `30^(@)`B. `45^(@)`C. `60^(@)`D. `90^(@)`

Answer» Correct Answer - A
`vec(A) . Vec(B) = AB cos theta = 2 sqrt(3)`
`|vec(A) xx vec(B)| = AB sin theta = 2`
`tan theta = (2)/( 2 sqrt(3)) = (1)/(sqrt(3)) rArr theta = 30^(@)`
Discussion
528.

A vector `hati+sqrt(3)hatj` rotates about its tail through an angle `60^(@)` in clockwise direction then the new vector isA. `hati+sqrt(3)hatj`B. `3hati-4hatj`C. `2hatj`D. `2hati`

Answer» Correct Answer - D
w.r.t. x-axis initially `theta=60^(@),|vecA|=2` units
on rotation `theta^(1)=0^(@),|vecA|^(1)=2` units
Discussion
529.

If `hatn` is a unit vector in the direction of the vector `vecA`, them :A. `hatn = vecA//|vecA|`B. `hatn = vecA|vecA|`C. `hatn = |vecA|//vecA`D. `hatn = vecn xx vecn`

Answer» Correct Answer - A
`hatn=(vecA)/(|vecA|)=("vector")/("magntiude")`
Discussion
530.

If `hatn` is a unit vector in the direction of the vector `vecA`, them :A. `A =vecA//hatn`B. `hatn = vecA//A`C. `1 = |vecA|//A`D. `vecA = Ahatn`

Answer» Correct Answer - B::C::D
Unit vector is defined as `hatn= (vecA)/(A)`.
Discussion
531.

The angle between `(vecA xx vec B)` and `(vecB xx vecA )` is :A. zeroB. `pi`C. `pi//4`D. `2pi//3`

Answer» Correct Answer - B
`(vecA xx vecB) and (vecB xx vecA)` have same magnitude but opposite direction
So, angle between them is `pi`.
Discussion
532.

If `vecP= vecQ` then which of the following is NOT correct?A. `hatP=hatQ`B. `|vecP|=|vecQ|`C. `PhatQ=QhatP`D. `vecP+vecQ=hatP+hatQ`

Answer» Correct Answer - D
Discussion
533.

`vecA=2hati+4hatj+4hatk` and `vecB=4hati+2hatj-4hatk` are two vectors. The angle between them will beA. `0^(@)`B. `45^(@)`C. `60^(@)`D. `90^(@)`

Answer» Correct Answer - D
Discussion
534.

O is apoint on the gournd chosen as origin. A boby first suffers a displacement of `10sqrt(2)`m North-East, next 10 m north and finally `10sqrt(2)` North-West. How far it is from the origin.A. 30n northB. 30 m southC. 30 m westD. 30 m east

Answer» Correct Answer - A
`vecs_(1)=10hati+10hatj,vecs_(2)=10hatj, vecs_(3)=-10hati+10hatj`
`vecs=vecs_(1)+vecs_(2)+vecs_(3)`
Discussion
535.

A particle moves towards east with velocity `5m//s`. After `10 seconds` its direction changes towards north with same Velocity. The average acceleration of the particle isA. ZeroB. `1/(sqrt(2))m//s^(2)N-W`C. `1/(sqrt(2))m//s^(2)N-E`D. `1/(sqrt(2))m//s^(2)S-W`

Answer» Correct Answer - B
Discussion
536.

Find the magnitudes of the unknown forces if the sum of all forces is zero (figure)

Answer» Correct Answer - `10/3; 25/3`
Sum of all given forces =0
`:.xhat(i)-10hat(j)-15cos53^(@)+15sin53^(@)hat(j)`
`-ycos37^(@)hat(i)-ysin37^(@)hat(j)=0`
`xhat(i)-15xx3/5hat(i)-y*4/5hat(i)=0`..(i)
`-10hat(j)+15xx4/5hat(j)-y*3/5hat(j)=0`..(ii)
`x-4/5y-9=0` or `x-4/5y=9`
`-10+12-3/5y=0` or `3/5y=2` or `y=10/3`
and `x=9+4/5xx10/3=25/3`
Discussion
537.

There are two force vectors, one of `5 N` and other of `12 N` at what angle the two vectors be added to get resultant vector of `17 N, 7 N` and `13 N` respectively.A. `0^(@), 180^(@)` and `90^(@)`B. `0^(@), 90^(@)` and `180^(@)`C. `0^(@), 90^(@)` ad `90^(@)`D. `180^(@), 0^(@)` and `90^(@)`

Answer» Correct Answer - A
Discussion
538.

If `AxxB=BxxA` then the angle between `A` and `B` isA. `pi//2`B. `pi//3`C. `pi`D. `pi//4`

Answer» Correct Answer - C
Discussion
539.

The angle between the vector `vecA` and `vecB` is `theta`. The value of the triple product `vecA.(vecBxxvecA)` isA. `A^(2)B`B. ZeroC. `A^(2)Bsintheta`D. `A^(2)costheta`

Answer» Correct Answer - B
Discussion
540.

Magnitudes of four pairs of displacement vectors are given. Which pair of displacment vectors, under vector addition fails to gives a resultant vectore of magnitude 3 cm ?A. 2 cm, 7 cmB. 1 cm,4 cmC. 2 cm, 3 cmD. 2 cm, 4 cm

Answer» Correct Answer - A
`R_("max") = a +b`
`R_("min") = |a -b|`
Discussion
541.

The resultant of two vectors `vec(A)` and `vec(B)` is perpendicular to the vector `vec(A)` and its magnitudes is equal to half of the magnitudes of vector `vec(B)` (figure). The angle between `vec(A)` and `vec(B)` is A. `120^(@)`B. `150^(@)`C. `135^(@)`D. None of these

Answer» Correct Answer - B
`cos beta=R/B=1/2rArr beta= 60^(@)`
Angle between `vec(A)` and `vec(B)= 90^(@)+beta=150^(@)`
Discussion
542.

The ratio of maximum and minimum magnitudes of the resultant of two vectors `vec(a)` and `vec(b)` is 3:1. Now, `|vec(a)|` is equal toA. `|vec(b)|`B. `2|vec(b)|`C. `3|vec(b)|`D. `4|vec(b)|`

Answer» Correct Answer - B
`(a+b)/(a-b)=3/1` or `3a-3b=a+b`
`2a=4b` or `a=2b`
Discussion
543.

Minimum numbar of vectors of unequal magnitudes which can give zore resultant areA. twoB. threeC. fourD. more than four

Answer» Correct Answer - B
Discussion
544.

Assertion: The minimum number of non-coplanar Vectors whose sum can be zero, is four Reason: The resultant of two vectors of unequal magnitude can be zero.A. If both assetion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is falseD. If the assertion and reason both are false.

Answer» Correct Answer - C
Discussion
545.

Assertion: The cross product of a vector with itself is a null vector. Reason: The cross-product of two vectors results in a vector quantity.A. If both assetion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is falseD. If the assertion and reason both are false.

Answer» Correct Answer - B
Discussion
546.

The vector sum of two vectors of magnitudes 10 units and 15 units can never beA. 28unitsB. 22unitsC. 18unitsD. 8units

Answer» Correct Answer - A
`vecP.vecQ` are two vectors:
`P+QgeRgeP-Q`
Discussion
547.

Can two vectors of different magitudes be combind to give zero resultant ?A. Yes, when the 2 vectors are same in magnitude and directionB. NoC. Yes, when the 2 vectors are same in magnitude but opposite in senseD. Yes, when the 2 vectors are same in magnitude making an angle of `(2pi)/3` with each other

Answer» Correct Answer - C
Discussion
548.

If for two vectors `hat(A)` and `hat(B)`,sum `(vec(A)+vec(B))` is perpendicular to the diffrence `(vec(A)-vec(B))`. Find the ratio of their magnitude.

Answer» `(vec(A)+vec(B))` is perpendiucar to `(vec(A)-vec(B))`
Thus `(vec(A)+vec(B)).(vec(A)-vec(B))=0`
or `A^(2)+vec(B).vec(A)-vec(A).vec(B)-B^(2)=0` Because of commutative property of dot produce `vec(A).vec(B)=vec(B).vec(A)`
`:. A^(2)-B^(2)` or `A=B`
Thus, the ratio of magnitudes `A//B=1`
Discussion
549.

The sum and diffrence of two perpendicular vector of equal length areA. Perpendicular to each other and of equal length.B. Perpendicular to each other and of different lengthC. Of equal length and have an obtuse angle between themD. Of equal length and have an acute angle between them

Answer» Correct Answer - A
Given `|vec(A)|=|vec(B)|` or `A=B`
Sum `vec(R )=vec(A)+vec(B) rArr |vec(R )|=sqrt(A^(2)+B^(2))=sqrt(2A)`
Difference `vec(S)=vec(A)-vec(B) rArr |vec(S)|=sqrt(A^(2)+B^(2))=sqrt(2A)`
`alpha_(1)=45^(@), alpha_(2)=45^(@)`
Hence, `vec(R )` and `vec(S)` will be perpendicular and also of equal lengths.
Discussion
550.

A force if `15N` acts on a box as shown in (figure).What are the horizontal component and vertical components of force?

Answer» Horizontal component`F_(x)=15cos60^(@)=7.5N`
Vertical component `F_(y)=15sin60^(@)=12.99N`.
Discussion