Let a, b be two non collinear unit vectors and x = a - ( a · b) b and y = a × b . Now consider the following statements: Statement-I: | x | = | y | Statement-II: | y | = | x | + | x · b |

Question

Let $\vec{a}, \vec{b}$ be two non collinear unit vectors and $\vec{x} = \vec{a} - (\vec{a} \cdot \vec{b}) \vec{b}$ and $\vec{y} = \vec{a} \times \vec{b}$ . Now consider the following statements:

Statement-I: $| \vec{x} |$ $=$ $| \vec{y} |$

Statement-II: $| \vec{y} |$ $=$ $| \vec{x} |$ $+$ $| \vec{x} \cdot \vec{b} |$

Smart Achievers · Accepted Answer

(3)

$Let \vec{a} \cdot \vec{b} = 0, Now \vec{x} = \vec{a} - (\vec{a} \cdot \vec{b}) \vec{b}$

$\Rightarrow | \vec{x} |^{2} = {\vec{a}}^{2} + {(\vec{a} \cdot \vec{b})}^{2} b^{2} - 2 {(\vec{a} \cdot \vec{b})}^{2}$

$\Rightarrow | \vec{x} |^{2} = 1 + \cos^{2} θ - 2 \cos^{2} θ \Rightarrow$ $| \vec{x} |$ $= \sin θ$

$Also \vec{y} = \vec{a} \times \vec{b}$ $\Rightarrow$ $| \vec{y} |$ $=$ $| \vec{a} \times \vec{b} |$ $= \sin θ$

$∴$ $| \vec{x} |$ $=$ $| \vec{y} |$

$Also, \vec{x} \cdot \vec{b} = \vec{a} \cdot \vec{b} - (\vec{a} \cdot \vec{b}) {\vec{b}}^{2} = 0$

$∴$ $| \vec{x} \cdot \vec{b} |$ $= 0$

Solution

1 Statement-I is true but Statement-II is false

2 Both Statement-I and Statement-II are false

3 Both Statement-I and Statement-II are true

4 Statement-I is false but Statement-II is true

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