Question 25. If |\vec{a}|=\sqrt{26} , |\vec{b}|= 7 and |\vec{a}\times\vec{b}|=35 , find \vec{a}.\vec{b}
Solution:
We know that,
=>
\vec{a}\times\vec{b} = |\vec{a}||\vec{b}|\sin\theta\hat{n} =>
|\vec{a}\times\vec{b} |= |\vec{a}||\vec{b}|\sin\theta\|hat{n}| =>
35 = \sqrt{26}\times7|\sin\theta|\times1 =>
\sin\theta = \dfrac{35}{7\sqrt{26}} =>
\sin\theta = \dfrac{5}{\sqrt{26}} As
\cos^2\theta + \sin^2\theta =1 ,=>
\cos\theta = \sqrt{1-\sin^2\theta} =>
\cos\theta = \sqrt{1-(\dfrac{5}{\sqrt{26}})^2} =>
\cos\theta = \sqrt{1-\dfrac{25}{26}} =>
\cos\theta = \dfrac{1}{\sqrt{26}} Thus,
=>
\vec{a}.\vec{b} = |\vec{a}||\vec{b}|\cos\theta =>
\vec{a}.\vec{b} = 7\sqrt{26}\times\dfrac{1}{\sqrt{26}} =>
\vec{a}.\vec{b} = 7
Question 26. Find the area of the triangle formed by O, A, B when \vec{OA} = \hat{i}+2\hat{j}+3\hat{k} , \vec{OB} = -3\hat{i}-2\hat{j}+\hat{k}
Solution:
The area of a triangle whose adjacent sides are given by
\vec{a} and\vec{b} is\dfrac{1}{2}|\vec{a}\times\vec{b}| =>
\vec{OA}\times\vec{OB} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\a_1 & a_2 & a_3\\b_1 & b_2 & b_3\end{vmatrix} =>
\vec{OA}\times\vec{OB} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1 & 2 & 3\\-3 & -2 & 1\end{vmatrix} =>
\vec{OA}\times\vec{OB} = \hat{i}[2+6]-\hat{j}[1+9]+\hat{k}[-2+6] =>
\vec{OA}\times\vec{OB} = 8\hat{i}-10\hat{j}+4\hat{k} => Area =
\dfrac{1}{2} |\vec{OA}\times\vec{OB}| => Area =
\dfrac{1}{2}\sqrt{8^2+(-10^2)+4^2} => Area =
\dfrac{1}{2}\sqrt{64+100+16} => Area =
\dfrac{1}{2}\times6\sqrt{45} => Area =
3\sqrt{5} square units.
Question 27. Let \vec{a}=\hat{i}+4\hat{j}+2\hat{k} , \vec{b}=3\hat{i}-2\hat{j}+7\hat{k} and \vec{c} = 2\hat{i}-\hat{j}+4\hat{k} . Find a vector \vec{d} which is perpendicular to both \vec{a} and \vec{b} and \vec{c}.\vec{d} = 15
Solution:
Given that
\vec{d} is perpendicular to both\vec{a} and\vec{b} .=>
\vec{d}.\vec{a} =0 ..........(1)=>
\vec{d}.\vec{b} =0 ..........(2)Also,
=>
\vec{c}.\vec{d} = 15 .......(3)Let
\vec{d} = d_1\hat{i}+d_2\hat{j}+d_3\hat{k} From eq(1),
=> d1 + 4d2 + 2d3 = 0
From eq(2),
=> 3d1 - 2d2 + 7d3 = 0
From eq(3),
=> 2d1 - d2 + 4d3 = 15
On solving the 3 equations we get,
d1 = 160/3, d2 = -5/3, and d3 = -70/3,
=>
\vec{d} = \dfrac {1}{3}(160\hat{i}-5\hat{j}-70\hat{k})
Question 28. Find a unit vector perpendicular to each of the vectors \vec{a}+\vec{b} and \vec{a}-\vec{b} , where \vec{a}=3\hat{i}+2\hat{j}+2\hat{k} and \vec{b} = \hat{i}+2\hat{j}-2\hat{k} .
Solution:
Given that,
\vec{a}=3\hat{i}+2\hat{j}+2\hat{k} and\vec{b} = \hat{i}+2\hat{j}-2\hat{k} Let
\vec{c} = \vec{a}+\vec{b} =>
\vec{c} = (3\hat{i}+2\hat{j}+2\hat{k})+ (\hat{i}+2\hat{j}-2\hat{k}) =>
\vec{c} = \hat{i}[3+1] +\hat{j}[2+2] +\hat{k}[2-2] =>
\vec{c} = 4\hat{i} +4\hat{j} Let
\vec{d} = \vec{a}-\vec{b} =>
\vec{d} = (3\hat{i}+2\hat{j}+2\hat{k})-(\hat{i}+2\hat{j}-2\hat{k}) =>
\vec{d} = \hat{i}[3-1] +\hat{j}[2-2] +\hat{k}[2+2] =>
\vec{d} = 2\hat{i} +4\hat{k} A vector perpendicular to both
\vec{c} and\vec{d} is,=>
\vec{c}\times\vec{d} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\c_1 & c_2 & c_3\\d_1 & d_2 & d_3\end{vmatrix} =>
\vec{c}\times\vec{d} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\4 & 4 & 0\\2 & 0 & 4\end{vmatrix} =>
\vec{c}\times\vec{d} = \hat{i}[16-0]-\hat{j}[16-0]+\hat{k}[0-8] =>
\vec{c}\times\vec{d} = 16\hat{i}-16\hat{j}-8\hat{k} To find the unit vector,
=>
\hat{p} = \dfrac{\vec{c}\times\vec{d}}{|\vec{c}\times\vec{d}|} =>
\hat{p} = \dfrac{1}{\sqrt{16^2+(-16)^2+(-8)^2}}(16\hat{i}-16\hat{j}-8\hat{k}) =>
\hat{p} = \dfrac{1}{\sqrt{256+256+64}}(16\hat{i}-16\hat{j}-8\hat{k}) =>
\hat{p} = \dfrac{1}{24}(16\hat{i}-16\hat{j}-8\hat{k}) =>
\hat{p} = \dfrac{1}{3}(2\hat{i}-2\hat{j}-\hat{k})
Question 29. Using vectors, find the area of the triangle with the vertices A(2, 3, 5), B(3, 5, 8), and C(2, 7, 8).
Solution:
Given, A(2, 3, 5), B(3, 5, 8), and C(2, 7, 8)
Let,
=>
\vec{a} = A = 2\hat{i}+3\hat{j}+5\hat{k} =>
\vec{b} = B = 3\hat{i}+5\hat{j}+8\hat{k} =>
\vec{c} = C = 2\hat{2}+7\hat{j}+8\hat{k} Then,
=>
\vec{AB} = \vec{b}-\vec{a} =>
\vec{AB} = (3\hat{i}+5\hat{j}+8\hat{k})-(2\hat{i}+3\hat{j}+5\hat{k}) =>
\vec{AB} = \hat{i}[3-2]+\hat{j}[5-3]+\hat{k}[8-5] =>
\vec{AB} = \hat{i}+2\hat{j}+3\hat{k} =>
\vec{AC} = \vec{c}-\vec{a} =>
\vec{AC} = (2\hat{2}+7\hat{j}+8\hat{k})-(2\hat{i}+3\hat{j}+5\hat{k}) =>
\vec{AC} = \hat{i}[2-2]+\hat{j}[7-3]+\hat{k}[8-5] =>
\vec{AC} = 4\hat{j}+3\hat{k} The area of a triangle whose adjacent sides are given by
\vec{a} and\vec{b} is\dfrac{1}{2}|\vec{a}\times\vec{b}| =>
\vec{AB}\times\vec{AC}= \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1 & 2& 3\\0 & 4 & 3\end{vmatrix} =>
\vec{AB}\times\vec{AC} = \hat{i}[6-12]-\hat{j}[3-0]+\hat{k}[4-0] =>
\vec{AB}\times\vec{AC} = -6\hat{i}-3\hat{j}+4\hat{k} => Area =
\dfrac{1}{2}|\vec{AB}\times\vec{AC}| => Area =
\dfrac{1}{2}\sqrt{(-6)^2+(-3)^2+4^2} => Area = √61/2
Question 30. If \vec{a}=2\hat{i}-3\hat{j}+\hat{k} , \vec{b}=-\hat{i}+\hat{k} , \vec{c}=2\hat{j}-\hat{k} are three vectors, find the area of the parallelogram having diagonals (\vec{a}+\vec{b}) and (\vec{b}+\vec{c}) .
Solution:
Given,
\vec{a}=2\hat{i}-3\hat{j}+\hat{k} ,\vec{b}=-\hat{i}+\hat{k} ,\vec{c}=2\hat{j}-\hat{k} Let,
=>
\vec{c} = (\vec{a}+\vec{b}) =>
\vec{c} = (2\hat{i}-3\hat{j}+\hat{k})+(-\hat{i}+\hat{k}) =>
\vec{c} = (2-1)\hat{i}+(-3)\hat{j}+(1+1)\hat{k} =>
\vec{c} = \hat{i}-3\hat{j}+2\hat{k} =>
\vec{d} = (\vec{b}+\vec{c}) =>
\vec{d} = (-\hat{i}+\hat{k})+(2\hat{j}-\hat{k}) =>
\vec{d} = -\hat{i}+2\hat{j} The area of the parallelogram having diagonals
\vec{c} and\vec{d} is\dfrac{1}{2}|\vec{c}\times\vec{d}| =>
\vec{c}\times\vec{d} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1 & -3& 2\\-1 & 2 & 0\end{vmatrix} =>
\vec{c}\times\vec{d} = \hat{i}[0-4] -\hat{j}[0+2]+\hat{k}[2-3] =>
\vec{c}\times\vec{d} = -4\hat{i}-2\hat{j}-\hat{k} => Area =
\dfrac{1}{2}|\vec{c}\times\vec{d}| => Area =
\dfrac{1}{2}\sqrt{(-4)^2+(-2)^2+(-1)^2} => Area =
\dfrac{1}{2}\sqrt{21} => Area = √21/2
Question 31. The two adjacent sides of a parallelogram are 2\hat{i}-4\hat{j}+5\hat{k} and \hat{i}-2\hat{j}-3\hat{k} . Find the unit vector parallel to one of its diagonals. Also, find its area.
Solution:
Given a parallelogram ABCD and its 2 sides AB and BC.
By triangle law of addition,
=>
\vec{AC} = \vec{AB}+\vec{BC} =>
\vec{AC} = (2\hat{i}-4\hat{j}+5\hat{k})+(\hat{i}-2\hat{j}-3\hat{k}) =>
\vec{AC} = \hat{i}[2+1] +\hat{j}[-4-2]+\hat{k}[5-3] =>
\vec{AC} = 3\hat{i}-6\hat{j}+2\hat{k} Unit vector is,
=>
\hat{p} = \dfrac{\vec{AC}}{|\vec{AC}|} =>
\hat{p} = \dfrac{1}{\sqrt{3^2+(-6)^2+2^2}}(3\hat{i}-6\hat{j}+2\hat{k}) =>
\hat{p} = \dfrac{1}{\sqrt{49}}(3\hat{i}-6\hat{j}+2\hat{k}) =>
\hat{p} = \dfrac{1}{7}(3\hat{i}-6\hat{j}+2\hat{k}) Area of a parallelogram whose adjacent sides are given is
|\vec{a}\times\vec{b}| =>
|\vec{AB}\times\vec{BC}| = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2 & -4 & 5\\1 & -2 & -3\end{vmatrix} =>
|\vec{AB}\times\vec{BC}| = \hat{i}[12+10]-\hat{j}[-6-5]+\hat{k}[-4+4] =>
|\vec{AB}\times\vec{BC}| = 22\hat{i}+11\hat{j} Thus area is,
=> Area =
|22\hat{i}+11\hat{j}| => Area =
\sqrt{22^2+11^2} => Area =
\sqrt{605} => Area = 11 √5 square units
Question 32. If either \vec{a}=0 or \vec{b}=0 , then \vec{a}\times\vec{b}=\vec{0} . Is the converse true? Justify with example.
Solution:
Let us take two parallel non-zero vectors
\vec{a} and\vec{b} =>
\vec{a}\times\vec{b} = \vec{0} For example,
\vec{a} = \hat{i} and\vec{b}=2\hat{i} =>
\vec{a}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1 & 0 & 0\\2 & 0 & 0\end{vmatrix} =>
\vec{a}\times\vec{b} = 0 But,
=>
|\vec{a}| = \sqrt{1^2} =1 =>
|\vec{b}| = \sqrt{2^2} =2 Hence the converse may not be true.
Question 33. If \vec{a} = a_1\hat{i}+a_2\hat{j}+a_3\hat{k} , \vec{b} = b_1\hat{i}+b_2\hat{j}+b_3\hat{k} and \vec{c} = c_1\hat{i}+c_2\hat{j}+c_3\hat{k} , then verify that \vec{a}\times(\vec{b}\times\vec{c})=\vec{a}\times\vec{b}+\vec{a}\times\vec{c} .
Solution:
Given,
\vec{a} = a_1\hat{i}+a_2\hat{j}+a_3\hat{k} ,\vec{b} = b_1\hat{i}+b_2\hat{j}+b_3\hat{k} and\vec{c} = c_1\hat{i}+c_2\hat{j}+c_3\hat{k} =>
(\vec{b}+\vec{c}) = (b_1+c_1)\hat{i}+(b_2+c_2)\hat{j}+(b_3+c_3)\hat{k} =>
\vec{a}\times(\vec{b}\times\vec{c}) = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\a_1 & a_2 & a_3\\(b_1+c_1) & (b_2+c_2) & (b_3+c_3)\end{vmatrix} =>
\vec{a}\times(\vec{b}\times\vec{c}) = \hat{i}[a_2(b_3+c_3)-a_3(b_2+c_2)]-\hat{j}[a_1(b_3+c_3)-a_3(b_1+c_1)]+\hat{k}[a_1(b_2+c_2)-a_2(b_1+c_1)] =>
\vec{a}\times(\vec{b}\times\vec{c}) = \hat{i}[a_2b_3+a_2c_3-a_3b_2-a_3c_2]+\hat{j}[-a_1b_3-a_1c_3+a_3b_1+a_3c_1]+\hat{k}[a_1b_2+a_1c_2-a_2b_1-a_2c_1] .....eq(1)Now,
=>
\vec{a}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\a_1 & a_2 & a_3\\b_1 & b_2 & b_3\end{vmatrix} =>
\vec{a}\times\vec{b} = \hat{i}[a_2b_3-b_2a_3]-\hat{j}[a_1b_3-b_1a_3]+\hat{k}[a_1b_2-b_1a_2] And,
=>
\vec{a}\times\vec{c} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\a_1 & a_2 & a_3\\c_1 & c_2 & c_3\end{vmatrix} =>
\vec{a}\times\vec{c} = \hat{i}[a_2b_3-c_2a_3]-\hat{j}[a_1c_3-c_1a_3]+\hat{k}[a_1c_2-c_1a_2] Thus,
=>
\vec{a}\times\vec{b}+\vec{a}\times\vec{c} = (\hat{i}[a_2b_3-b_2a_3]-\hat{j}[a_1b_3-b_1a_3]+\hat{k}[a_1b_2-b_1a_2]) + (\hat{i}[a_2b_3-c_2a_3]-\hat{j}[a_1c_3-c_1a_3]+\hat{k}[a_1c_2-c_1a_2]) =>
\vec{a}\times\vec{b}+\vec{a}\times\vec{c} = \hat{i}[a_2b_3+a_2c_3-a_3b_2-a_3c_2]+\hat{j}[-a_1b_3-a_1c_3+a_3b_1+a_3c_1]+\hat{k}[a_1b_2+a_1c_2-a_2b_1-a_2c_1] ...eq(2)Thus eq(1) = eq(2)
Hence proved.
Question 34(i). Using vectors find the area of the triangle with the vertices A(1, 1, 2), B(2, 3, 5), and C(1, 5, 5).
Solution:
Given, A(1, 1, 2), B(2, 3, 5), and C(1, 5, 5)
=>
\vec{a} = A = \hat{i}+\hat{j}+2\hat{k} =>
\vec{b} = B = 2\hat{i}+3\hat{j}+5\hat{k} =>
\vec{c} = C = \hat{i}+5\hat{j}+5\hat{k} Now 2 sides of the triangle are given by,
=>
\vec{AB} = \vec{b}-\vec{a} =>
\vec{AB} = (2\hat{i}+3\hat{j}+5\hat{k})-(\hat{i}+\hat{j}+2\hat{k}) =>
\vec{AB} = \hat{i}[2-1] +\hat{j}[3-1]+\hat{j}[5-2] =>
\vec{AB} = \hat{i}+2\hat{j}+3\hat{k} =>
\vec{AC} = \vec{c}-\vec{a} =>
\vec{AC} = (\hat{i}+5\hat{j}+5\hat{k})-(\hat{i}+\hat{j}+2\hat{k}) =>
\vec{AC} = \hat{i}[1-1] +\hat{j}[5-1]+\hat{j}[5-2] =>
\vec{AC} = 4\hat{j}+3\hat{k} Area of the triangle whose adjacent sides are given is
\dfrac{1}{2}|\vec{a}\times\vec{b}| =>
\vec{AB}\times\vec{AC} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1 & 2 & 3\\0 & 4 & 3\end{vmatrix} =>
\vec{AB}\times\vec{AC} = \hat{i}[6-12]-\hat{j}[3-0]+\hat{k}[4-0] =>
\vec{AB}\times\vec{AC} = -6\hat{i}-3\hat{j}+4\hat{k} Thus area of the triangle is,
=> Area =
\dfrac{1}{2}\sqrt{(-6)^2+(-3)^2+4^2} => Area =
\dfrac{1}{2}\sqrt{36+9+16} => Area = √61/2
Question 34(ii). Using vectors find the area of the triangle with the vertices A(1, 2, 3), B(2, -1, 4), and C(4, 5, -1).
Solution:
Given, A(1, 2, 3), B(2, -1, 4), and C(4, 5, -1)
=>
\vec{a} = A = \hat{i}+2\hat{j}+3\hat{k} =>
\vec{b} = B = 2\hat{i}-1\hat{j}+4\hat{k} =>
\vec{c} = C = 4\hat{i}+5\hat{j}-1\hat{k} Now 2 sides of the triangle are given by,
=>
\vec{AB} = \vec{b}-\vec{a} =>
\vec{AB} = (2\hat{i}-1\hat{j}+4\hat{k})-(4\hat{i}+5\hat{j}-1\hat{k}) =>
\vec{AB} = \hat{i}[2-1] +\hat{j}[-1-2]+\hat{j}[4-3] =>
\vec{AB} = \hat{i}-3\hat{j}+\hat{k} =>
\vec{AC} = \vec{c}-\vec{a} =>
\vec{AC} = (4\hat{i}+5\hat{j}-1\hat{k})-(\hat{i}+2\hat{j}+3\hat{k}) =>
\vec{AC} = \hat{i}[4-1] +\hat{j}[5-2]+\hat{j}[-1-3] =>
\vec{AC} = 3\hat{i}+3\hat{j}-4\hat{k} Area of the triangle whose adjacent sides are given is
\dfrac{1}{2}|\vec{a}\times\vec{b}| =>
\vec{AB}\times\vec{AC} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1 & -3 & 1\\3 & 3 & -4\end{vmatrix} =>
\vec{AB}\times\vec{AC} = \hat{i}[12-3]-\hat{j}[-4-3]+\hat{k}[3+9] =>
\vec{AB}\times\vec{AC} = 9\hat{i}+7\hat{j}+12\hat{k} Thus area of the triangle is,
=> Area =
\dfrac{1}{2}\sqrt{(9)^2+(7)^2+12^2} => Area =
\dfrac{1}{2}\sqrt{81+49+144} => Area = √274/2
Question 35. Find all the vectors of magnitude 10\sqrt{3} that are perpendicular to the plane of \hat{i}+2\hat{j}+\hat{k} and -\hat{i}+3\hat{j}+4\hat{k} .
Solution:
Given,
\vec{a} = \hat{i}+2\hat{j}+\hat{k} and\vec{b}=\hat{i}+3\hat{j}+4\hat{k} A vector perpendicular to both
\vec{a} and\vec{b} is,=>
\vec{a}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1 & 2 & 1\\-1 & 3 & 4\end{vmatrix} =>
\vec{a}\times\vec{b} = \hat{i}[8-3]-\hat{j}[4+1]+\hat{k}[3+2] =>
\vec{a}\times\vec{b} = 5\hat{i}-5\hat{j}+5\hat{k} Unit vector is,
=>
\hat{p} = \dfrac{\vec{a}\times\vec{b}}{|\vec{a}\times\vec{b}|} =>
\hat{p} = \dfrac{1}{\sqrt{5^2+(-5)^2+5^2}}(5\hat{i}-5\hat{j}+5\hat{k}) =>
\hat{p} = \dfrac{1}{5\sqrt{3}}(5\hat{i}-5\hat{j}+5\hat{k}) =>
\hat{p} = \dfrac{1}{\sqrt{3}}(\hat{i}-\hat{j}+\hat{k}) Now vectors of magnitude
10\sqrt{3} are given by,=>
10\sqrt{3}\hat{p} = \pm10\sqrt{3}\times \dfrac{1}{\sqrt{3}}(\hat{i}-\hat{j}+\hat{k}) => Required vectors,
\pm10(\hat{i}-\hat{j}+\hat{k})
Question 36. The adjacent sides of a parallelogram are 2\hat{i}-4\hat{j}-5\hat{k} and 2\hat{i}+2\hat{j}+3\hat{k} . Find the 2 unit vectors parallel to its diagonals. Also, find its area of the parallelogram.
Solution:
Given,
\vec{AB}=2\hat{i}-4\hat{j}-5\hat{k} and\vec{BC} = 2\hat{i}+2\hat{j}+3\hat{k} =>
\vec{AC} = \vec{AB} + \vec{BC} =>
\vec{AC} = (2\hat{i}-4\hat{j}-5\hat{k})+(2\hat{i}+2\hat{j}+3\hat{k}) =>
\vec{AC} = 4\hat{i}-\hat{j}-2\hat{k} Unit vector is,
=>
\hat{p} = \dfrac{\vec{AC}}{|\vec{AC}|} =>
\hat{p} = \dfrac{1}{\sqrt{4^2+(-1)^2+(-2)^2}}(4\hat{i}-\hat{j}-2\hat{k}) =>
\hat{p} = \dfrac{1}{\sqrt{21}}(4\hat{i}-\hat{j}-2\hat{k}) Area is given by
|\vec{AB}\times\vec{BC}| ,
Summary
The cross product of two vectors a and b is a vector perpendicular to both a and b. Key points include:
1. Magnitude: |a × b| = |a||b|sin θ, where θ is the angle between a and b.
2. Direction: Right-hand rule determines the direction of a × b.
3. Properties: a × b = -(b × a), a × a = 0, distributive over addition.
4. In component form: a × b = (a₂b₃ - a₃b₂)i + (a₃b₁ - a₁b₃)j + (a₁b₂ - a₂b₁)k.
Practice Problems
1. If a = 2i + 3j - k and b = i - 2j + 4k, find a × b.
2. Find a unit vector perpendicular to both a = i + 2j - k and b = 2i - j + 2k.
3. Prove that (a × b) · (b × c) = (a · b)(b · c) - b²(a · c).
4. If a = 3i - 2j + k, b = i + j - k, and c = 2i + 3j - 2k, find the volume of the parallelepiped formed by these vectors.
5. Show that (a × b) × c = (a · c)b - (b · c)a.
6. Find the area of the triangle formed by the points A(1, 2, 3), B(2, 3, 1), and C(3, 1, 2).
7. Prove that |a × (b × c)| = |b||c|sin θ, where θ is the angle between b and c.
8. If |a| = 3, |b| = 4, and |a × b| = 6, find the angle between a and b.
9. Find λ such that the vectors a = 2i - j + 3k, b = i + 2j - k, and c = λi + 3j + 2k are coplanar.
10. Prove that (a × b) · (c × d) = (a · c)(b · d) - (a · d)(b · c).