Question 1(i). Evaluate the following
Solution:
= =
= 1 + 1 + 1
= 3
Question 1(ii). Evaluate the following
Solution:
= =
= 2 – 1 – 2
= -1
Question 2(i). Find
Solution:
= = 2(-1 – 0) + 3(-1 + 3)
= -2 + 6
= 4
Question 2(ii). Find
Solution:
= = 1(1 + 1) + 2(2 + 0) + 3(2 – 0)
= 2 + 4 + 6
= 12
Question 3(i). Find the volume of the parallelepiped whose coterminous edges are represented by vector
Solution:
Volume of a parallelepiped whose adjacent edges are
is equal to
= = 2(4 – 1) – 3(2 + 3) + 4(-1 – 6)
= 6 – 15 – 28
= -9 – 28
= -37
So, Volume of parallelepiped is | -37 | = 37 cubic unit.
Question 3(ii). Find the volume of the parallelepiped whose coterminous edges are represented by vector
Solution:
Volume of a parallelepiped whose adjacent edges
are equal to
= = 2(-4 – 1) + 3(-2 + 3) + 4(-1 – 6)
= -10 + 3 – 28
= -10 – 25
= -35
So, Volume of parallelepiped = | -35 | = 35 cubic unit.
Question 3(iii). Find the volume of the parallelepiped whose coterminous edges are represented by vector
Solution:
Let a = 11
, b = 2 , c = 13 Volume of a parallelepiped whose adjacent edges are
is equal to
= = 11(26 – 0) + 0 + 0
= 286
Volume of a parallelepiped = | 286| = 286 cubic units.
Question 3(iv). Find the volume of the parallelepiped whose coterminous edges are represented by vector
Solution:
Let
Volume of a parallelepiped whose adjacent edges
are equal to
= = 1(1 – 2) – 1(-1 – 1) + 1(2 + 1)
= -1 + 2 + 3
= 4
Volume of a parallelepiped = |4| = 4 cubic units.
Question 4(i). Show of the following triads of vector is coplanar :
Solution:
As we know that three vectors
are coplanar if their = 0.
= = 1(10 – 42) – 2(15 – 35) – 1(18 – 10)
= -32 + 40 – 8
= 0
So, the given vectors are coplanar.
Question 4(ii). Show of the following triads of vector is coplanar :
Solution:
As we know that three vectors
are coplanar if their = 0.
= = -4(12 + 3) + 6(-3 + 24) – 2(1 + 32)
= -60 + 126 – 66
= 0
So, the given vectors are coplanar.
Question 4(iii). Show of the following triads of vector is coplanar :
Solution:
As we know that three vectors
are coplanar if their = 0.
= = 1(15 – 12) + 2(-10 + 4) + 3(6 – 3)
= 3 – 12 + 9
= 0
So, the given vectors are coplanar.
Question 5(i). Find the value of λ so that the following vector is coplanar:
Solution:
As we know that three vectors
are coplanar if their = 0.
= = 1(λ -1) + 1(2λ + λ) + 1(-2 – λ)
= λ – 1 + 3λ – 2 -λ
3 = 3λ
1 = λ
So, the value of λ is 1
Question 5(ii). Find the value of λ so that the following vector is coplanar:
Solution:
As we know that three vectors
are coplanar if their = 0.
= = 2(10 + 3 λ) + 1(5 + 3 λ) + 1(λ – 2 λ)
= 20 + 6 λ + 5 + 3 λ – λ
-25 = 8 λ
λ = – 25 / 8
So, the value of λ is -25/8
Question 5(iii). Find the value of λ so that the following vector is coplanar:
Solution:
Given:
As we know that three vectors
are coplanar if their = 0.
= = 1(2λ – 2) – 2(6 – 1) – 3(6 – λ)
= 2λ – 2 -12 + 2 -18 + 3λ
= 5λ – 30
30 = 5λ
λ = 6
So, the value of the λ is 6
Question 5(iv). Find the value of λ so that the following vector is coplanar:
Solution:
Given:
So, to prove that these points are coplanar, we have to prove that
= 0
= = 1(0 + 5) – 3(0 – 5λ) + 0
= 5 + 15λ
-5 = 15λ
λ = – 1 / 3
Question 6. Show that the four points having position vectors
Solution:
Let us considered
OA =
OB =
OC =
OD =
AB = OB – OA =
AC = OC – OA =
CD = OD – OC =
AD = OD – OA =
So, to prove that these points are coplanar, we have to prove that
= 16(-160 – 24) + 25(-160 + 8) – 4(-144 + 64) ≠ 0
Hence, proved that the points are not coplanar.
Question 7. Show that the points A (-1, 4, -3), B(3, 2, -5), C(-3, 8, -5), and D(-3, 2, 1) are coplanar
Solution:
Given:
A = (-1, 4, -3)
B = (3, 2, -5)
C = (-3, 8, -5)
D = (-3, 2, 1)
=
=
= So, to prove that these points are coplanar, we have to prove that
Thus,
= 4[16 – 4] + 2[-8 -4] – 2[4 + 8]
= 48 – 24 – 24 = 0
Hence, proved.
Question 8. Show that four points whose position vectors are
Solution:
Let us considered
OA =
OB =
OC =
OD =
Thus,
AB = OB – OA =
AC = OC – OA =
AD = OD – OA =
If the vectors AB, AC and AD are coplanar then the four points are coplanar
On simplifying, we get
= 10(70 + 12) + 12(-30 – 24) – 4(-6 + 28)
= 820 – 648 – 88
= 84 ≠ 0
So, the points are not coplanar.
Question 9. Find the value of λ for which the four points with position vectors
Solution:
Let us considered:
Position vector of A =
Position vector of B =
Position vector of C =
Position vector of D =
If the given vectors
are coplanar, then the four points are coplanar
=
=
= On simplifying, we get
4(50 – 25) – 6(15 + 20) + (λ + 1)(15 + 40) = 0
100 – 210 + 55 + 55λ = 0
55λ = 55
λ = 1
So, when the value of λ = 1, the given points are coplanar.
Question 10. Prove that
Solution:
Given:
One solving the given equation we get
=
=
= 6 [ a b c ] – 6 [ a b c ]
= 0
Hence proved
Question 11.
Solution:
In the given triangle ABC,
If
= AB
= BC
= AC Then,
is perpendicular to the plane of the given triangle ABC
is perpendicular to the plane of the given triangle ABC
is perpendicular to the plane of the given triangle ABC Hence, proved that
is a vector perpendicular to the plane of the given triangle ABC.
Question 12(i). Let
Solution:
Given:
are coplanar only if = 0
0 – 1(C3) + 1(2) = 0
C3 = 2
So, when the value C3 = 2, then these points are coplanar.
Question 12(ii). Let
Solution:
Given:
are coplanar only if = 0 So,
0 – 1 + 1 (C1) = 0
C1 = 1
Hence, prove that no value of C1 can make these points coplanar
Question 13. Find λ for which the points A (3, 2, 1), B (4, λ, 5), C (4, 2, -2), and D (6, 5, -1) are coplanar
Solution:
Let us considered:
Position vector of OA =
Position vector of OB =
Position vector of OC =
Position vector of OD =
If the vectors AB, AC, and AD are coplanar, then the four points are coplanar
AB =
AC =
AD =
On simplifying, we get
1(9) – (λ – 2)(-2 + 9) + 4(3 – 0) = 0
9 – 7 λ + 14 + 12 = 0
7 λ = 35
λ = 5
Hence, the value of λ is 5. So the coplanar points are, A(3, 2, 1), B(4, 5, 5), C(4, 2, -2), and D(6, 5, -1)