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Question: The non zero vectors \(a,b\) and \(c\) are related by \(a=8b,c=-7b\), then angle between \(a\) and \...

The non zero vectors a,ba,b and cc are related by a=8b,c=−7ba=8b,c=-7b, then angle between aa and cc is?
1. π\pi
2. 0
3. π4\dfrac{\pi }{4}
4. π2\dfrac{\pi }{2}

Explanation

Solution

For solving this question you should know about the characteristics of triple vectors. The product of the triple vectors is given by a×(b×c)=b(c.a)−c(a.b)a\times \left( b\times c \right)=b\left( c.a \right)-c\left( a.b \right) and compare it with the given parameters. First write all these in vector form and then use the formula cos⁡x=a.c∣a∣∣c∣\cos x=\dfrac{a.c}{\left| a \right|\left| c \right|} to find the angle.

Complete step-by-step solution:
According to the question it is asked to find the angle between aa and cc if the non-zero vectors a,ba,b and cc are related by a=8b,c=−7ba=8b,c=-7b. Consider the problem, it is given that,
a→=8b→\overrightarrow{a}=8\overrightarrow{b} and c→=−7b→\overrightarrow{c}=-7\overrightarrow{b}
If we write these in the vector form, then it can be written as follows,
b=b1i^+b2j^+b3k^b={{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k}
And a→=8b→\overrightarrow{a}=8\overrightarrow{b}
a→=8(b1i^+b2j^+b3k^)\overrightarrow{a}=8\left( {{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k} \right)
And c→=−7b→\overrightarrow{c}=-7\overrightarrow{b}
c→=−7(b1i^+b2j^+b3k^)\overrightarrow{c}=-7\left( {{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k} \right)
As we know that the product of three non-zero factors is a×(b×c)=b(c.a)−c(a.b)a\times \left( b\times c \right)=b\left( c.a \right)-c\left( a.b \right), but here we have to find the angle between aa and cc, so we will use the formula cos⁡x=a.c∣a∣∣c∣\cos x=\dfrac{a.c}{\left| a \right|\left| c \right|}. Applying the formula we will get as follows,
⇒cos⁡x=−56(b12+b22+b32)7×8b12+b22+b32.b12+b22+b32 =−56(b12+b22+b32)56(b12+b22+b32) =−5656=−1 \begin{aligned} & \Rightarrow \cos x=\dfrac{-56\left( {{b}_{1}}^{2}+{{b}_{2}}^{2}+{{b}_{3}}^{2} \right)}{7\times 8\sqrt{{{b}_{1}}^{2}+{{b}_{2}}^{2}+{{b}_{3}}^{2}}.\sqrt{{{b}_{1}}^{2}+{{b}_{2}}^{2}+{{b}_{3}}^{2}}} \\\ & =\dfrac{-56\left( {{b}_{1}}^{2}+{{b}_{2}}^{2}+{{b}_{3}}^{2} \right)}{56\left( {{b}_{1}}^{2}+{{b}_{2}}^{2}+{{b}_{3}}^{2} \right)} \\\ & =\dfrac{-56}{56}=-1 \\\ \end{aligned}
Therefore cos⁡x=−1\cos x=-1 and you get the value of xx as follows,
x=cos⁡−1(−1) x=π \begin{aligned} & x={{\cos }^{-1}}\left( -1 \right) \\\ & x=\pi \\\ \end{aligned}
Hence the correct answer is option 1.

Note: While solving this type of question you have to keep in mind that the angle between two vectors can be 0 to π\pi . And it is always calculated by the dot product of both of these. And first you have to find the modulus value of that and then use it to find the angle.