The lines r=(6-6s)a+(4s-4)b+(4-8s)c  and r=(2t-1)a+(4t-2)b-(2t+3)c intersects at

1.  4c

2.  -4c

3.  3c

4.  -2c

4

-4c

Explanation :
No Explanation available for this question

I : If cos α + cos β + cos γ = 3 then sin α + sin β + sin γ = 0 II: If sin α + sin β + sin γ =

1.  Only I is true

2.  Only II is true

3.  Both I & II are true

4.  Neither I nor II are true

4

Both I & II are true

Explanation :
No Explanation available for this question

I: If tan A + tan B = P and Cot A + Cot B = q then cot (A + B) = (q-p) / pq II : If 2 tan A + cot A = tan B then cot A +2 tan (A –B) = 0

1.  Only I is true

2.  Only II is true

3.  Both I & II are true

4.  Neither I nor II are true

4

Both I & II are true

Explanation :
No Explanation available for this question

If the position vectors of A, B, C, D are 3i+2j+k, 4i+5j+5k, 4i+2j-2k, 6i+5j-k respectively then the position vector of the point of intersection of

1.  2i+j-3k

2.  2i-j+3k

3.  2i+j+3k

4.  2i-j-3k

4

2i-j-3k

Explanation :
No Explanation available for this question

If P is appoint on the line passing through A(3i+j-k) and parallel to 2i-j+2k such that AP=15 then position vector of P is

1.  13i-4j+9k

2.  3i+4j+9k

3.  13i-4j-9k

4.  13i-4j-9k

4

13i-4j+9k

Explanation :
No Explanation available for this question

The vector equation of the plane passing through the point 2i+2j-3k and parallel to the vectors 3i+3j-5k, i+2j+k is

1.  r=s(2i+j-k)+t(i+2j+2k)

2.  r=2i+2j-3k+s(3i+3j-5k)+t(i+2j+k)

3.  r=(i+2j+3k)+s(-2i+3j+k)+t(2i-3j+4k)

4.  none

4

r=2i+2j-3k+s(3i+3j-5k)+t(i+2j+k)

Explanation :
No Explanation available for this question

The vector equation of the plane passing through the point (1, -2, -3) and parallel to the vectors (2, -1, 3), (2, 3, -6) is

1.  r=(i-2j-3k)+s(2i-j+3k)+t(2i+3j-6k)

2.  r=(1-s-t)(i-2j-3k)+s(2i-j+3k)+t(2i+3j-6k)

3.  r=(i-2j-3k)+s(4j-9k)

4.  r=(34j-9k)+s(i-2j-3k)

4

r=(i-2j-3k)+s(2i-j+3k)+t(2i+3j-6k)

Explanation :
No Explanation available for this question

The vector equation of the plane passing through the point (1, -2, 5), (0, -5, -1), (-3, 5, 0) is

1.  r=(1-s-t)(i-2j+5k)+s(-5j-k)+t(-3i+5j), s, t are scalars

2.  r=(1-s-t)(i+2j+3k)+s(3i+2j+k)+t(2i+j+3k), s, t are scalars

3.  r=(1-s-t)(2i+j+k)+s(i-j-k)+t(-i+j+2k), s, t are scalars

4.  r=(1-s-t)(i-2j+5k)+s(-5j-k)+t(-3i+5j), s, t are scalars

4

r=(1-s-t)(i-2j+5k)+s(-5j-k)+t(-3i+5j), s, t are scalars

Explanation :
No Explanation available for this question

The vector equation of the plane passing through the point i+2j+5k, -5j+k, -3i+5j is

1.  r=(1-s-t)(i-2j+5k)+s(-5j+k)+t(-3i+5j), s, t are scalars

2.  r=(1-s-t)(i+2j+3k)+s(3i+2j+k)+t(2i+j+3k), s, t are scalars

3.  r=(1-s-t)(2i+j+k)+s(i-j-k)+t(-i+j+2k), s, t are scalars

4.  r=(1-s-t)(i-2j+5k)+s(-5j-k)+t(-3i+5j), s, t are scalars

4

r=(1-s-t)(i-2j+5k)+s(-5j+k)+t(-3i+5j), s, t are scalars

Explanation :
No Explanation available for this question

The points 2a-b+3c, -a+2b-4c, 12a-b-3c, 6a+2b-c lie in a plane. Then the vector equation of the plane is

1.  r=(1-s-t)(2a-b+3c)+s(-a+2b-4c )+t(-12a-b-3c), s, t are scalars

2.  r=(1+s-t) (2a-b+3c)+s(-a+2b-4c )+t(-12a-b+3c),, s, t are scalars

3.  r=(1-s+t) (2a-b+3c)+s(-a+2b-4c )+t(-12a-b-3c),, s, t are scalars

4.   r=(1-s-t) (2a-b+3c)+s(-a+2b-4c )+t(-12a-b-3c), s, t are scalars

4