Vector doubts..

for the 1st qstn....

A = B + C

C = A - B

A = ai + bj + ck
B = di + 3j + 4k
C = 3i + j - 2k

3i + j - 2k = ai + bj + ck - di - 3j - 4k

3i + j - 2k = (a - d)i + (b - 3)j +(c - 4)k

Comparing both sides, we get....

a - d = 3

b = 4

c = 2

Nw, i think u get it.... solve a and d wid the help of area.....

plz help in the 2nd one...

area will be 12 |AB x AC| , may be this can help

Is the answer for the second question i-2j-2k?

2) i can think of some ways . but m nt sure whther they will work out.

u can consider a plane .. in which the initial and final vectors lie. (one vector is given vector .. other is the vector obtained when the vector is rotated till it coincides with x-axis) ..

u can hence find the eqn of plane .

u can also find the equation of the line which contains the posn vector and lies in the plane..

hence u can find the eqn of line which is perp. to this line and lying in the plane ..
now length of vector remains const. so ull get two coordinates. choose appropriate one

PS: even here ..we r nt asked to transform coordinates abt such dangerous axes [3] ..so rest assured that dis ques. wont cum .. at max. what can come is rotation in a plane that is parallel to xy or yz or zx plane.. they cant possibly give anything more calculative than dis

A vector OA=i + 2j + 2k turns through a right angle, passing through the positive x-axis on the way. The vector in the new position is ________.

dirn of normal = OA x i = <0, 2, -2>

so eqn of plane : y-z = 0

eqn of line containing P : (x-1)/1 = (y-2)/2 = (z-2)/2

Let dc of new line be <a,b,c>

then a.1 + b.2 + c.2 = 0 ... (1)
a²+b²+c² = 1 .. (2)
a.0 + b.2 + c.(-2) = 0 ... (3)

=> <a,b,c> = <-43√2, 13√2, 13√2>

So, dr of new line are <-4,1,1>

Now let new line has eqn : x = at, y = bt, z = ct
where a,b,c are dr of new line
and
x²+y²+z² = 9 (length of vec)

18 t² = 9
=> t = ±1/√2

=> x,y,z = ±(-4√2, 1√2, 1√2)

but x coord of rotated vector should be positive

=> rotated point is (4√2, -1√2, -1√2)

hope its correct [1]