# Mathematics Questions and Answers – Product of Two Vectors-1

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This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Product of Two Vectors-1”.

1. Find the scalar product of the vectors $$\vec{a}=2\hat{i}+5\hat{j}$$ and $$\vec{b}=6\hat{i}-7\hat{j}$$.
a) -32
b) -23
c) 32
d) 23

Explanation: If $$\vec{a} \,and \,\vec{b}$$ are two vectors, where a1, a2 are the components of vector $$\vec{a} \,and \,b_1, \,b_2$$ are the components of vector $$\vec{b}$$, then the scalar product is given by
$$\vec{a}.\vec{b}=a_1 \,b_1+a_1 \,b_2$$
∴$$(2\hat{i}+5\hat{j}).(6\hat{i}-7\hat{j})$$=2(6)+5(-7)=12-35=-23.

2. Find the angle between the two vectors $$\vec{a}$$ and $$\vec{b}$$ with magnitude $$\sqrt{3}$$ and $$\sqrt{2}$$ respectively and $$\vec{a.} \,\vec{b}=3\sqrt{2}$$.
a) $$cos^{-1}⁡\frac{1}{\sqrt{3}}$$
b) $$cos^{-1}⁡\sqrt{3}$$
c) $$cos^{-1}⁡\frac{3}{\sqrt{2}}$$
d) $$cos^{-1}⁡\frac{2}{\sqrt{3}}$$

Explanation: Given that, $$|\vec{a}|=\sqrt{3} \,and \,|\vec{b}|=\sqrt{2}$$
Also, $$\vec{a.} \vec{b}=3\sqrt{2}$$
The angle between two vectors is given by
$$cos⁡θ=\frac{|\vec{a}|.|\vec{b}|}{\vec{a}.\vec{b}}$$
∴$$cos⁡θ=\frac{\sqrt{3}.\sqrt{2}}{3\sqrt{2}}=\frac{1}{\sqrt{3}}$$
∴$$θ=cos^{-1}⁡\frac{1}{\sqrt{3}}$$.

3. Find the projection of vector $$\vec{a}=8\hat{i}-\hat{j}+6\hat{k}$$ on vector $$\vec{b}= 4\hat{i}+3\hat{j}$$.
a) $$\sqrt{\frac{29}{5}}$$
b) $$\frac{29}{\sqrt{5}}$$
c) $$\frac{\sqrt{29}}{5}$$
d) $$\frac{29}{5}$$

Explanation: The projection of a vector $$\vec{a}$$ on vector $$\vec{b}$$ is given by
$$\frac{1}{|\vec{b}|} (\vec{a}.\vec{b})$$
$$|\vec{b}|=\sqrt{4^2+3^2}=\sqrt{16+9}$$=5
$$\vec{a}.\vec{b}$$=8(4)-1(3)+0=32-3=29
The projection of vector $$8\hat{i}-\hat{j}+6\hat{k}$$ on vector $$4\hat{i}+3\hat{j}$$ will be
$$\frac{1}{|\vec{b}|} (\vec{a}.\vec{b})=\frac{1}{5} (29)=\frac{29}{5}$$
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4. Find $$|\vec{a}+\vec{b}|$$, if $$|\vec{a}|=3 \,and \,|\vec{b}|=4 \,and \,\vec{a}.\vec{b}=6$$.
a) 34
b) $$\sqrt{37}$$
c) 13
d) $$\sqrt{23}$$

Explanation: $$|\vec{a}+\vec{b}|^2=(\vec{a}+\vec{b}).(\vec{a}+\vec{b})$$
=$$\vec{a}.\vec{a}+\vec{a}.\vec{b}+\vec{b}.\vec{a}+\vec{b}.\vec{b}$$
=$$|\vec{a}|^2+2(\vec{a}.\vec{b})+|\vec{b}|^2$$
=(3)2+2(6)+(4)2
=9+12+16=37
∴$$|\vec{a}+\vec{b}|=\sqrt{37}$$

5. Find the angle between the vectors $$\vec{a}=\hat{i}-\hat{j}+2\hat{k} \,and \,\vec{b}=3\hat{i}+2\hat{j}+4\hat{k}$$.
a) $$cos^{-1}⁡\sqrt{\frac{58}{3}}$$
b) $$cos^{-1}⁡\frac{\sqrt{58}}{3}$$
c) $$cos^{-1}\frac{⁡58}{3\sqrt{3}}$$
d) $$cos^{-1}⁡\frac{\sqrt{58}}{3\sqrt{3}}$$

Explanation: The angle between the two vectors is given by
$$cos⁡θ=\frac{|\vec{a}|.|\vec{b}|}{\vec{a}.\vec{b}}$$
$$|\vec{a}|=\sqrt{1^2+(-1)^2+2^2}=\sqrt{1+1+4}=\sqrt{6}$$
$$|\vec{b}|=\sqrt{3^2+2^2+(-4)^2}=\sqrt{9+4+16}=\sqrt{29}$$
$$\vec{a}.\vec{b}$$=1(3)-1(2)+2(4)=9
∴$$cos⁡θ=\frac{\sqrt{6}.\sqrt{29}}{9}=\frac{\sqrt{58}}{3\sqrt{3}}$$
∴$$θ=cos^{-1}\frac{⁡\sqrt{58}}{3\sqrt{3}}$$

6. Find the angle between the vectors $$\vec{a}=-\hat{i}+\hat{j}-\hat{k}$$ and $$\vec{b}=\hat{i}-\hat{j}$$
a) $$cos^{-1}⁡-\frac{\sqrt{3}}{2}$$
b) $$cos^{-1}⁡-\frac{2}{\sqrt{3}}$$
c) $$cos^{-1}⁡-\sqrt{2}$$
d) $$cos^{-1}⁡-\sqrt{\frac{3}{2}}$$

Explanation: The angle between two vectors is given by
$$cos⁡θ=\frac{|\vec{a}|.|\vec{b}|}{\vec{a}.\vec{b}}$$
$$|\vec{a}|=\sqrt{(-1)^2+(1)^2+(-1)^2}=\sqrt{3}$$
$$|\vec{b}|=\sqrt{(1)^2+(-1)^2}=\sqrt{2}$$
$$\vec{a}.\vec{b}$$=(-1)(1)+1(-1)+0=-2
$$cos⁡θ=\frac{\sqrt{3}.\sqrt{2}}{-2}=-\sqrt{\frac{3}{2}}$$
∴$$θ=cos^{-1}⁡-\sqrt{\frac{3}{2}}$$

7. If two non-zero vectors $$\vec{a} \,and \, \vec{b}$$ are perpendicular to each other then their scalar product is zero.
a) True
b) False

Explanation: The given statement is true. If the angle between two vectors is $$\frac{π}{2}$$ i.e. they are perpendicular to each other, their scalar product will be zero.

8. Find the angle between the two vectors $$\vec{a} \,and \, \vec{b}$$ with magnitude 2 and $$\sqrt{3}$$ respectively and $$\vec{a.} \, \vec{b}$$=4.
a) $$\frac{π}{3}$$
b) $$\frac{π}{6}$$
c) $$cos^{-1}⁡\frac{\sqrt{2}}{3}$$
d) $$cos^{-1}⁡\frac{2}{\sqrt{3}}$$

Explanation: Given that, $$|\vec{a}|=2 \,and \,|\vec{b}|=\sqrt{3}$$
Also, $$\vec{a.} \,\vec{b}=4$$
The angle between two vectors is given by
$$cos⁡θ=\frac{|\vec{a}|.|\vec{b}|}{\vec{a}.\vec{b}}$$
∴$$cos⁡θ=\frac{2.\sqrt{3}}{4}=\frac{\sqrt{3}}{2}$$
∴$$θ=cos^{-1}⁡\frac{\sqrt{3}}{2}=\frac{π}{6}$$.

9. Find the scalar product of the vectors $$\vec{a}=6\hat{i}-7\hat{j}+5\hat{k} \,and \,\vec{b}=6\hat{i}-7\hat{k}$$
a) 1
b) 8
c) 6
d) 3

Explanation: If $$\vec{a} \,and \,\vec{b}$$ are two vectors, where $$a_1, a_2, a_3$$ are the components of vector $$\vec{a} \,and \,b_1, b_2, b_3$$ are the components of vector $$\vec{b}$$, then the scalar product is given by
$$\vec{a}.\vec{b}=a_1 b_1+a_1 b_2+a_3 b_3$$
$$(6\hat{i}-7\hat{j}+5\hat{k}).(6\hat{i}-7\hat{k})$$=6(6)-7(0)+5(-7)=36-35=1.

10. Find the projection of vector $$\vec{b}=2\hat{i}+2\sqrt{2} \,\hat{j}-2\hat{k}$$ on the vector $$\vec{a}=\hat{i}-\hat{j}-\sqrt{2} \,\hat{k}$$.
a) 2
b) $$\sqrt{2}$$
c) 1
d) $$2\sqrt{2}$$

Explanation: The projection of vector $$\vec{b}$$ on the vector $$\vec{b}$$ is given by $$\frac{1}{|\vec{a}|} (\vec{a}.\vec{b})$$
$$|\vec{a}|=\sqrt{(1)^2+(-1)^2+(-\sqrt{2})^2}=\sqrt{1+1+2}=\sqrt{4}$$=2
Also, $$\vec{a}.\vec{b}=2(1)+2\sqrt{2} \,(-1)-2(-\sqrt{2})=2-2\sqrt{2}+2\sqrt{2}$$=2
Therefore, the projection of vector $$\hat{i}-\hat{j}-\sqrt{2} \,\hat{k}$$ on the vector $$\vec{b}=2\hat{i}+2\sqrt{2}\hat{j}-2\hat{k}$$ is
$$\frac{1}{|\vec{a}|} (\vec{a}.\vec{b})=\frac{1}{2}$$ (2)=1.

Sanfoundry Global Education & Learning Series – Mathematics – Class 12.

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