第 12 类 RD Sharma 解决方案 – 第 26 章标量三倍积 – 练习 26.1
问题 1(i)。评估以下内容
解决方案:
=
=
= 1 + 1 + 1
= 3
问题 1(ii)。评估以下内容
解决方案:
=
=
= 2 – 1 – 2
= -1
问题 2(i)。找 , 什么时候
解决方案:
=
= 2(-1 – 0) + 3(-1 + 3)
= -2 + 6
= 4
问题 2(ii)。找 , 什么时候
解决方案:
=
= 1(1 + 1) + 2(2 + 0) + 3(2 – 0)
= 2 + 4 + 6
= 12
问题 3(i)。求平行六面体的体积,其相邻边由向量表示
解决方案:
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.
问题 3(ii)。求平行六面体的体积,其相邻边由向量表示
解决方案:
Volume of a parallelepiped whose adjacent edges are is 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.
问题 3(iii)。求平行六面体的体积,其相邻边由向量表示
解决方案:
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.
问题 3(iv)。求平行六面体的体积,其相邻边由向量表示
解决方案:
Let
Volume of a parallelepiped whose adjacent edges are is equal to
=
= 1(1 – 2) – 1(-1 – 1) + 1(2 + 1)
= -1 + 2 + 3
= 4
Volume of a parallelepiped = |4| = 4 cubic units.
问题 4(i)。下列三元组向量的显示是共面的:
解决方案:
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.
问题 4(ii)。下列三元组向量的显示是共面的:
解决方案:
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.
问题 4(iii)。下列三元组向量的显示是共面的:
解决方案:
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.
问题 5(i)。求 λ 的值,使以下向量共面:
解决方案:
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
问题 5(ii)。求 λ 的值,使以下向量共面:
解决方案:
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
问题 5(iii)。求 λ 的值,使以下向量共面:
解决方案:
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
问题 5(iv)。求 λ 的值,使以下向量共面:
解决方案:
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
问题 6. 证明具有位置向量的四个点不共面。
解决方案:
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.
问题 7. 证明点 A (-1, 4, -3), B(3, 2, -5), C(-3, 8, -5) 和 D(-3, 2, 1) 是共面
解决方案:
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.
问题 8. 证明四个点的位置向量是
解决方案:
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.
问题 9. 求具有位置向量的四个点的 λ 值是共面的
解决方案:
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.
问题 10. 证明
解决方案:
Given:
One solving the given equation we get
=
=
= 6 [ a b c ] – 6 [ a b c ]
= 0
Hence proved
问题 11。 分别是点 A、B 和 C 的位置向量,证明是垂直于三角形 ABC 平面的向量。
解决方案:
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.
问题 12(i)。让 .然后,如果 c 1 = 1 和 c 2 = 2,找到 c 3使得共面。
解决方案:
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.
问题 12(ii)。让和 .然后,如果 c2 = -1 且 c3 =1,证明没有 c 1的值可以使共面
解决方案:
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
问题 13. 求点 A (3, 2, 1)、B (4, λ, 5)、C (4, 2, -2) 和 D (6, 5, -1) 共面的 λ
解决方案:
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)