00:06Question says, the expression 1 upon under root 2 i plus 1 upon under root 2 j is a a
00:16unit vector, b null vector, c vector of magnitude under root 2 d scalar.
00:27Suppose, the p vector is equal to 1 upon under root 2 i plus 1 upon under root 2 j.
00:43So, the
00:44magnitude of this vector will be equal to under root x square plus y square. So,
00:58magnitude of p vector will be obtained as the value of x is 1 upon under root 2 and
01:06the value of y is 1 upon under root 2 because the unit vector along x axis is i and
01:15the unit
01:16vector along y axis is j. So, the value of x is 1 upon under root 2 whole square because
01:26x square plus y square. So, the value of y is also 1 upon under root 2. So, whole square
01:36and
01:361 upon under root 2. So, the magnitude of vectors p will be obtained as the square of 1 is
01:491 and the square of under root 2 is 2 plus the square of 1 is 1 and the square
01:56of under root 2 is 2. So, by taking LCM,
02:01the LCM, the LCM of 2 is 2. So, it will be the table of 2 and 2 is 1
02:09times 1 and 1 plus the table of 2 and 2 is 1 times 1 and 1, it will be
02:161. So, it will be 1 plus 1, 2 divided by 2 means under root 2 divided by 2.
02:29So, the magnitude of p vector will be 1. So, the magnitude of p vector will be under root 1
02:46which is equal to 1.
02:46So, as we know that a vector having magnitude 1 or unity is known as unit vector. So, this is
02:59a unit vector.
03:05So, the correct option will be a unit vector because the magnitude of unit vector is always 1.
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