QuizSolver

Solved Objective Question on Vector Set 1

Posted on - 04-01-2017

IIT JEE

Question.1

then

(A)

(B)

(C)

(D) none of these

Solution

Question.2

The points with position vector 60i + 3j, 40i – 8j and ai –52j are collinear if

(A) a = -40

(B) a = 40

(C) a = 20

(D) none of these

Solution

The points are collinear
&⇒ l( 60i + 3j ) + m( 40 i- 8j) + n( ai- 52j) = 0 with l+m+ n = 0
&⇒ 60 l + 40 m + na = 0

3 l - 8m - 52n = 0

l + m + n = 0

For non-zero set ( l, m, n) , = 0
&⇒ a = -40

Question.3

are three vectors respectively given by 2i +k, i + j+ k, and
4 i – 3j + 7k. Then vector , which satisfies the relation , is

(A) 2 i – 5j + 2k

(B) –i + 4j + 2k

(C) -i – 8j + 2k

(D) none of these

Solution

We have

&⇒
&⇒

&⇒ ( 2 +1)= 3– ( 8+7)
&⇒ =

Question.4

If the lines intersect ( t and s are scalars) then.

(A)

(B)

(C)

(D) none of these

Solution

For the point of intersection of the lines

Question.5

The number of vectors of unit length perpendicular to vector a º ( 1, 1, 0) and b º ( 0, 1, 1) is

(A) 1

(B) 2

(C) 3

(D) 4

Solution

The vector of unit length perpendicular to the given vectors

= .

Hence there are two such vectors . .

Question.6

Let º 2 i – j + k , º i+ 2j – k and = i + j – 2k be three vectors. A vector in the plane of and whose projection on is of magnitude is

(A) 2i + 3j – 3k

(B) 2i + 3j + 3k

(C) – 2i - j + 5k

(D) 2i + j + 5k

Solution

Let be a vector in the plane of b and c

&⇒ = ( i + 2j – k) + m ( i + j – 2k)

Its projection on = =[2 +2 m - 2 - m -1 - 2m ] =

&⇒ =
&⇒ -( 1+m) = ± 2
&⇒ m = 1, -3

&⇒ R º 2 i + 3j – 3k and - 2i – j + 5k .

Question.7

The scalar equals

(A) 0

(B)

(C)

(D) none of these

Solution

=

&⇒

Question.8

Let be two unit vectors such that is also a unit vector. Then the angle between is

(A) 30°

(B) 60°

(C) 90°

(D) 120°

Solution

Sinceis a unit vector, ().() =1

&⇒ + + 2=1
&⇒ 1 +1 + 2 =1
&⇒ = -1/2

Hence the angle q between is given by cosq = -1/2

&⇒ q =120°.

Question.9

If vectors axand xmake an acute angle with each other, for all x ∈ R, then a belongs to the interval

(A)

(B) ( 0, 1)

(C)

(D)

Solution

Since vectors make an acute angle with each other so their dot product must be positive. .

i.e. ax² – 10 ax + 6 > 0 " x ∈ R.

&⇒-ax² + 10ax – 6 < 0 " x ∈ R

&⇒-a < 0 and 100a²< 24a

Question.10

If are unit vectors, such that = 0, then the value of is

(A) 1

(B) -1

(C) -3/2

(D) None of these

Solution

DABC is an equilateral

\

Alternate

=

&⇒ = –

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