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Maths-

General

Easy

Question

$stack P Q with rightwards arrow on top comma stack Q R with rightwards arrow on top comma stack R S with rightwards arrow on top comma stack S T with rightwards arrow on top comma stack T U with rightwards arrow on top text and end text stack U P with rightwards arrow on top$ represent the sides of a regular hexagon
Statement I: $stack P Q with rightwards arrow on top cross times open parentheses stack R S with rightwards arrow on top plus stack S T with rightwards arrow on top close parentheses not equal to 0$
Statement II: $stack P Q with rightwards arrow on top cross times stack R S with rightwards arrow on top equals 0$ and $stack P Q with rightwards arrow on top cross times stack S T with rightwards arrow on top not equal to stack 0 with rightwards arrow on top$

Both Statement I and II are true and Statement II is a correct explanation for Statement I.
Both Statement I and II are true but Statement II is not a correct explanation for Statement I.
Statement I is true and Statement II is false
Statement I is false and Statement II is true

The correct answer is: Statement I is true and Statement II is false

We have
$stack P Q with rightwards arrow on top cross times open parentheses stack R S with rightwards arrow on top plus stack S T with rightwards arrow on top close parentheses equals stack P Q with rightwards arrow on top cross times stack R T with rightwards arrow on top not equal to 0$
Because $stack P Q with rightwards arrow on top$ and $stack R T with bar on top$ are not parallel. Statement I is true. Now
$stack P Q with rightwards arrow on top cross times stack R S with rightwards arrow on top not equal to stack 0 with rightwards arrow on top$
because they are not parallel vectors. Also
$stack P Q with rightwards arrow on top cross times stack S T with rightwards arrow on top equals stack 0 with rightwards arrow on top text because end text stack P Q with rightwards arrow on top text and end text stack S T with rightwards arrow on top$ are parallel
Hence Statement II is false.

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