Question

Three vectors $7i-11j+k$, $5i+3j-2k$ and $12i-8j-k$ forms?

• A. An equilateral triangle
• B. An isosceles triangle
• C. A right-angled triangle
• D. None of these

Answer 1

The correct option is C

A right-angled triangle

Explanation for the correct option:

Step 1: Find the modulus of the given vectors.

In the question, Three vectors $7i-11j+k$, $5i+3j-2k$ and $12i-8j-k$ are given.

Find the modulus of $7i-11j+k$.

$$\begin{gathered} \left|7i-11j+k\right|=\sqrt{7^2+{\left(-11\right)}^2+1^2} \\ \Rightarrow \left|7i-11j+k\right|=\sqrt{49+121+1} \\ \Rightarrow \left|7i-11j+k\right|=\sqrt{171} \end{gathered}$$

Find the modulus of $5i+3j-2k$.

$$\begin{gathered} \left|5i+3j-2k\right|=\sqrt{5^2+3^2+{(-2)}^2} \\ \Rightarrow \left|5i+3j-2k\right|=\sqrt{25+9+4} \\ \Rightarrow \left|5i+3j-2k\right|=\sqrt{38} \end{gathered}$$

Find the modulus of $12i-8j-k$.

$$\begin{gathered} \left|12i-8j-k\right|=\sqrt{{12}^2+{(-8)}^2+{(-1)}^2} \\ \Rightarrow \left|12i-8j-k\right|=\sqrt{144+64+1} \\ \Rightarrow \left|12i-8j-k\right|=\sqrt{209} \end{gathered}$$

Therefore, the modulus of $7i-11j+k$, $5i+3j-2k$ and $12i-8j-k$ are $\sqrt{171},\sqrt{38}$ and $\sqrt{209}$ respectively.

Step 2: Determine the type of triangle formed by the given vectors.

Find the square of the modulus of all the given vectors.

$$\begin{gathered} {\left|7i-11j+k\right|}^2=171...1 \\ {\left|5i+3j-2k\right|}^2=38...2 \\ {\left|12i-8j-k\right|}^2=209...3 \end{gathered}$$

It is clear from equation $1,2$ and $3$.

$$\begin{gathered} 171+38=209 \\ \Rightarrow {\left|7i-11j+k\right|}^2+{\left|5i+3j-2k\right|}^2={\left|12i-8j-k\right|}^2 \end{gathered}$$

Since the given vectors satisfy Pythagora's theorem.

Therefore, the given three vectors $7i-11j+k$, $5i+3j-2k$ and $12i-8j-k$ forms a right-angle triangle.

Hence, option $C$ is the correct option.