Mathematics Question Papers for Class 12

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

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=-\(\frac{7}{2}\).

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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
\(\vec{a}×\vec{b}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&4&-5\\3&-1&2\end{vmatrix}\)
=\(\hat{i}(8-5)-\hat{j}(4-(-15))+\hat{k}(-2-12)\)
=\(3\hat{i}-19\hat{j}-14\hat{k}\)

Check this: Class 12 - Mathematics Books | Class 12 - Physics MCQs

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: \((\vec{a}+\vec{b}).(7\vec{a}-6\vec{b})\)
=\(\vec{a}.7\vec{a}-\vec{a}.6\vec{b}+\vec{b}.7\vec{a}-6\vec{b}.\vec{b}\)
=\(7|\vec{a}|^2+\vec{a}.\vec{b}-6|\vec{b}|^2\)

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

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.

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