The sides of a parallelogram are given by the vectors -2-4-5- and -1-2-3- - nbsp-then the unit vector parallel to one of the diagonals is1-7-3 -6-2k-1-7-3 -6-2k-1-7-3 -6-2k-1-7-3 -6-2k-

The correct option is A 1/7(3 î+6ĵ 2k̂)Let a= (2,4, 5), b= (1,2,3). nbsp;Then the diagonals of the parallelogram are p=a+b, q=b a ⇒p= nbsp; (3,6, 2) , q= ( ...

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Standard XII

Mathematics

Addition of Vectors

The sides of ...

Question

The sides of a parallelogram are given by the vectors $(2, 4, - 5)$ and $(1, 2, 3)$ , then the unit vector parallel to one of the
diagonals is

\frac{1}{7} (3^i + 6^j - 2^k)

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\frac{1}{7} (3^i - 6^j + 2^k)

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\frac{1}{7} (- 3^i + 6^j - 2^k)

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\frac{1}{7} (3^i + 6^j + 2^k)

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Solution

The correct option is A $\frac{1}{7} (3^i + 6^j - 2^k)$
Let $\to a = (2, 4, - 5)$ , $\to b = (1, 2, 3)$ . Then the diagonals of the parallelogram are $\to p = \to a + \to b, \to q = \to b - \to a$
$\Rightarrow \to p = (3, 6, - 2), \to q = (- 1, - 2, 8)$
So, unit vectors along the diagonals are $\frac{1}{7} (3, 6, - 2)$ and $\frac{1}{\sqrt{69}} (- 1, - 2, 8)$

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Similar questions

Q. Show that each of the given three vectors is a unit vector:

\frac{1}{7} (2^i + 3^j + 6^k), \frac{1}{7} (3^i - 6^j + 2^k), \frac{1}{7} (6^i + 2^j - 3^k)

.
Also, show that they are mutually perpendicular to each other.

Q. The vector

\to a = 1 / 7 (2^i + 3^j + 6^k)

\to b = 1 / 7 (3^i - 6^j + 2^k)

\to c = 1 / 7 (6^i + 2^j - 3^k)

form

Q. Show that each of the three given vectors is a unit vectors. also Show that they are mutually

⊥

to each other.

\to a = \frac{1}{7} (2^i + 3^j + 6^k)

\to b = \frac{1}{7} (3^i - 6^j + 2^k)

\to c = \frac{1}{7} (6^i + 2^j - 3^k)

Q. The area (in sq. units) of the parallelogram whose diagonals are along the vectors

8^i - 6^j

and

3^i + 4^j - 12^k

, is.

Q. The area

(i n s q . u n i t s)

of the parallelogram whosed diagonals along the vectors

8^i - 6^j

and

3^i + 4^j - 12^k

, is:

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The sides of a parallelogram are given by the vectors (2,4,−5) and (1,2,3) , then the unit vector parallel to one of the diagonals is

The correct option is A 17(3^i+6^j−2^k)Let →a=(2,4,−5), →b=(1,2,3). Then the diagonals of the parallelogram are →p=→a+→b,→q=→b−→a ⇒→p=(3,6,−2),→q=(−1,−2,8) So, unit vectors along the diagonals are 17(3,6,−2) and 1√69(−1,−2,8)

The sides of a parallelogram are given by the vectors $(2, 4, - 5)$ and $(1, 2, 3)$ , then the unit vector parallel to one of the
diagonals is