Mathematics Questions and Answers – Product of Two Vectors-2

«
»

This set of Mathematics Question Papers for Class 12 focuses on “Product of Two Vectors-2”.

1. Evaluate the product \((2\vec{a}+5\vec{b}).(4\vec{a}-5\vec{b})\).
a) \(|\vec{a}|^2+2\vec{a}.\vec{b}-15|\vec{b}|^2\)
b) \(8|\vec{a}|^2+2\vec{a}.\vec{b}-15|\vec{b}|^2\)
c) \(8|\vec{a}|^2-4\vec{a}.\vec{b}-15|\vec{b}|^2\)
d) \(|\vec{a}|^2+\vec{a}.\vec{b}-5|\vec{b}|^2\)
View Answer

Answer: b
Explanation: To evaluate: \((2\vec{a}+5\vec{b}).(4\vec{a}-5\vec{b})\)
=\(2\vec{a}.4\vec{a}-2\vec{a}.5\vec{b}+3\vec{b}.4\vec{a}-3\vec{b}.5\vec{b}\)
=\(8|\vec{a}|^2+2\vec{a}.\vec{b}-15|\vec{b}|^2\)
advertisement

2. Find the magnitude of \(\vec{a}\) and \(\vec{b}\) which are having the same magnitude and such that the angle between them is 60° and their scalar product is \(\frac{1}{4}\).
a) \(|\vec{a}|=|\vec{b}|=\frac{1}{2√2}\)
b) \(|\vec{a}|=|\vec{b}|=\frac{1}{√2}\)
c) \(|\vec{a}|=|\vec{b}|=\frac{1}{2√3}\)
d) \(|\vec{a}|=|\vec{b}|=\frac{2}{√3}\)
View Answer

Answer: a
Explanation: Given that: a) \(|\vec{a}|=|\vec{b}|\)
b) θ=60°
c) \(\vec{a}.\vec{b}=\frac{1}{4}\)
∴\(|\vec{a}||\vec{b}| cos⁡θ=\frac{1}{4}\)
=\(|\vec{a}|^2 cos⁡60°=\frac{1}{4}\)
⇒\(|\vec{a}|^2=\frac{1}{4}.\frac{1}{2}\)
∴\(|\vec{a}|=|\vec{b}|=\frac{1}{2√2}\).

3. If \(\vec{a}=\hat{i}-\hat{j}+3\hat{k}, \,\vec{b}=5\hat{i}-2\hat{j}+\hat{k} \,and \,\vec{c}=\hat{i}-\hat{j}\) are such that \(\vec{a}+μ\vec{b}\) is perpendicular to \(\vec{c}\), then the value of μ.
a) \(\frac{7}{2}\)
b) –\(\frac{7}{2}\)
c) –\(\frac{3}{2}\)
d) \(\frac{7}{9}\)
View Answer

Answer: b
Explanation: Given that: \(\vec{a}=\hat{i}-\hat{j}+3\hat{k}, \,\vec{b}=5\hat{i}-2\hat{j}+\hat{k} \,and \,\vec{c}=\hat{i}-\hat{j}\)
Also given, \(\vec{a}+μ\vec{b}\) is perpendicular to \(\vec{c}\)
Therefore, \((\vec{a}+μ\vec{b}).\vec{c}=0\)
i.e. \((\hat{i}-\hat{j}+3\hat{k}+μ(5\hat{i}-2\hat{j}+\hat{k})).(\hat{i}-\hat{j})\)=0
\(((1+5μ) \,\hat{i}-(1+2μ) \,\hat{j}+(μ+3) \,\hat{k}).(\hat{i}-\hat{j})\)=0
1+5μ+1+2μ=0
μ=-\(\frac{7}{2}\).
advertisement
advertisement

4. Find the angle between \(\vec{a} \,and \,\vec{b}\) if \(|\vec{a}|=2,|\vec{b}|=\frac{1}{2√3}\) and \(\vec{a}×\vec{b}=\frac{1}{2}\).
a) \(\frac{2π}{3}\)
b) \(\frac{4π}{5}\)
c) \(\frac{π}{3}\)
d) \(\frac{π}{2}\)
View Answer

Answer: c
Explanation: Given that, \(|\vec{a}|=2, \,|\vec{b}|=\frac{1}{2√3}\) and \(\vec{a}×\vec{b}=\frac{1}{2}\)
We know that, \(\vec{a}×\vec{b}=\vec{a}.\vec{b} \,sin⁡θ\)
∴ \(sin⁡θ=\frac{(\vec{a}×\vec{b})}{|\vec{a}|.|\vec{b}|}\)
sin⁡θ=\(\frac{\frac{1}{2}}{2×\frac{1}{2√3}}=\frac{\sqrt{3}}{2}\)
θ=\(sin^{-1}⁡\frac{\sqrt{3}}{2}=\frac{π}{3}\)

5. Find the vector product of the vectors \(\vec{a}=2\hat{i}+4\hat{j}\) and \(\vec{b}=3\hat{i}-\hat{j}+2\hat{k}\).
a) \(\hat{i}-19\hat{j}-4\hat{k}\)
b) \(3\hat{i}+19\hat{j}-14\hat{k}\)
c) \(3\hat{i}-19\hat{j}-14\hat{k}\)
d) \(3\hat{i}+5\hat{j}+4\hat{k}\)
View Answer

Answer: c
Explanation: Given that, \(\vec{a}=2\hat{i}+4\hat{j}\) and \(\vec{b}=3\hat{i}-\hat{j}+2\hat{k}\)
Calculating the vector product, we get
\(\vec{a}×\vec{b}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&4&-5\\3&-1&2\end{vmatrix}\)
=\(\hat{i}(8-5)-\hat{j}(4-(-15))+\hat{k}(-2-12)\)
=\(3\hat{i}-19\hat{j}-14\hat{k}\)
advertisement

6. If \(\vec{a} \,and \,\vec{b}\) are two non-zero vectors then \((\vec{a}-\vec{b})×(\vec{a}+\vec{b})\)=_________
a) \(2(\vec{a}×\vec{b})\)
b) \((\vec{a}×\vec{b})\)
c) –\(4(\vec{a}×\vec{b})\)
d) \(3(\vec{a}×\vec{b})\)
View Answer

Answer: a
Explanation: Consider \((\vec{a}-\vec{b})×(\vec{a}+\vec{b})\)
=\((\vec{a}-\vec{b})×\vec{a}+(\vec{a}+\vec{b})×\vec{b}\)
=\(\vec{a}×\vec{a}-\vec{b}×\vec{a}+\vec{a}×\vec{b}-\vec{b}×\vec{b}\)
We know that, \(\vec{a}×\vec{a}=0,\vec{b}×\vec{b}=0 \,and \,\vec{a}×\vec{b}=-\vec{b}×\vec{a}\)
∴ \(\vec{a}×\vec{a}-\vec{b}×\vec{a}+\vec{a}×\vec{b}-\vec{b}×\vec{b}=0+2(\vec{a}×\vec{b})+0\)
Hence, \((\vec{a}-\vec{b})×(\vec{a}+\vec{b})=2(\vec{a}×\vec{b})\)

7. Find the product \((\vec{a}+\vec{b}).(7\vec{a}-6\vec{b})\).
a) \(2|\vec{a}|^2+6\vec{a}.\vec{b}-3|\vec{b}|^2\)
b) \(8|\vec{a}|^2+5\vec{a}.\vec{b}-5|\vec{b}|^2\)
c) \(2|\vec{a}|^2+6\vec{a}.\vec{b}-8|\vec{b}|^2\)
d) \(7|\vec{a}|^2+\vec{a}.\vec{b}-6|\vec{b}|^2\)
View Answer

Answer: d
Explanation: To evaluate: \((\vec{a}+\vec{b}).(7\vec{a}-6\vec{b})\)
=\(\vec{a}.7\vec{a}-\vec{a}.6\vec{b}+\vec{b}.7\vec{a}-6\vec{b}.\vec{b}\)
=\(7|\vec{a}|^2+\vec{a}.\vec{b}-6|\vec{b}|^2\)
advertisement

8. Find the vector product of the vectors \(\vec{a}=-\hat{j}+\hat{k}\) and \(\vec{b}=-\hat{i}-\hat{j}-\hat{k}\).
a) \(2\hat{i}-\hat{j}+\hat{k}\)
b) \(2\hat{i}-\hat{j}-4\hat{k}\)
c) \(\hat{i}+\hat{j}-\hat{k}\)
d) \(2\hat{i}-\hat{j}-\hat{k}\)
View Answer

Answer: d
Explanation: Given that, \(\vec{a}=-\hat{j}+\hat{k}\) and \(\vec{b}=-\hat{i}-\hat{j}-\hat{k}\)
Calculating the vector product, we get
\(\vec{a}×\vec{b}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\0&-1&1\\-1&-1&-1\end{vmatrix}\)
=\(\hat{i}(1-(-1))-\hat{j}(0-(-1))+\hat{k}(0-1)\)
=\(2\hat{i}-\hat{j}-\hat{k}\)

9. If \(\vec{a}=2\hat{i}+3\hat{j}+4\hat{k}\) and \(\vec{b}=4\hat{i}-2\hat{j}+3\hat{k}\). Find \(|\vec{a}×\vec{b}|\).
a) \(\sqrt{685}\)
b) \(\sqrt{645}\)
c) \(\sqrt{679}\)
d) \(\sqrt{689}\)
View Answer

Answer: b
Explanation: Given that, \(\vec{a}=2\hat{i}+3\hat{j}+4\hat{k}\) and \(\vec{b}=4\hat{i}-2\hat{j}+3\hat{k}\)
∴ \(\vec{a}×\vec{b}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&3&4\\4&-2&3\end{vmatrix}\)
=\(\hat{i}(9—8)-\hat{j}(6-16)+\hat{k}(-4-12)\)
=\(17\hat{i}+10\hat{j}-16\hat{k}\)
∴\(|\vec{a}×\vec{b}|=\sqrt{17^2+10^2+(-16)^2}\)
=\(\sqrt{289+100+256}\)
=\(\sqrt{645}\)
advertisement

10. Find the angle between the vectors if \(|\vec{a}|=|\vec{b}|=3\sqrt{2}\) and \(\vec{a}.\vec{b}=9\sqrt{3}\).
a) \(\frac{π}{6}\)
b) \(\frac{π}{5}\)
c) \(\frac{π}{3}\)
d) \(\frac{π}{2}\)
View Answer

Answer: a
Explanation: We know that, \(\vec{a}.\vec{b}=|\vec{a}|.|\vec{b}| \,cos⁡θ\)
Given that, \(|\vec{a}|=|\vec{b}|=3\sqrt{2} \,and \,\vec{a}.\vec{b}=9\sqrt{3}\)
\(cos⁡θ=\frac{\vec{a}.\vec{b}}{|\vec{a}|.|\vec{b}|}=\frac{9\sqrt{3}}{(3\sqrt{2})^2}=\frac{\sqrt{3}}{2}\)
\(θ=cos^{-1}\frac{\sqrt{3}}{2}=\frac{π}{6}\).

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

To practice Mathematics Question Papers for Class 12, here is complete set of 1000+ Multiple Choice Questions and Answers.

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!

advertisement
advertisement
Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter