# Mathematics Questions and Answers – Product of Two Vectors-2

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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) $$|\vec{a}|^2+\vec{a}.\vec{b}-5|\vec{b}|^2$$

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) $$|\vec{a}|=|\vec{b}|=\frac{2}{√3}$$

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) $$\frac{7}{9}$$

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) $$\frac{π}{2}$$

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) $$3\hat{i}+5\hat{j}+4\hat{k}$$

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

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) $$3(\vec{a}×\vec{b})$$

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) $$7|\vec{a}|^2+\vec{a}.\vec{b}-6|\vec{b}|^2$$

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) $$2\hat{i}-\hat{j}-\hat{k}$$

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) $$\sqrt{689}$$

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) $$\frac{π}{2}$$

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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