Mathematics Questions and Answers – Product of Two Vectors-1

«
»

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
View Answer

Answer: b
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.
advertisement

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) \(cos^{-1}⁡\frac{2}{\sqrt{3}}\)
View Answer

Answer: a
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) \(\frac{29}{5}\)
View Answer

Answer: d
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
\(\frac{1}{|\vec{b}|} (\vec{a}.\vec{b})=\frac{1}{5} (29)=\frac{29}{5}\)
Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now!
advertisement
advertisement

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) \(\sqrt{23}\)
View Answer

Answer: b
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) \(cos^{-1}⁡\frac{\sqrt{58}}{3\sqrt{3}}\)
View Answer

Answer: d
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
∴\(cos⁡θ=\frac{\sqrt{6}.\sqrt{29}}{9}=\frac{\sqrt{58}}{3\sqrt{3}}\)
∴\(θ=cos^{-1}\frac{⁡\sqrt{58}}{3\sqrt{3}}\)
advertisement

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) \(cos^{-1}⁡-\sqrt{\frac{3}{2}}\)
View Answer

Answer: d
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
View Answer

Answer: a
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.
advertisement

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) \(cos^{-1}⁡\frac{2}{\sqrt{3}}\)
View Answer

Answer: b
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
View Answer

Answer: a
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.
advertisement

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) \(2\sqrt{2}\)
View Answer

Answer: b
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.

To practice all areas of Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.

advertisement
advertisement
Subscribe to our Newsletters (Subject-wise). Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

Youtube | Telegram | LinkedIn | Instagram | Facebook | Twitter | Pinterest
Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

Subscribe to his free Masterclasses at Youtube & technical discussions at Telegram SanfoundryClasses.