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Addition of Vectors

Let ABCD be...

Question

Let $A B C D$ be a parallelogram. If $A B =^i + 3^j + 7^k, A D = 2^i + 3^j - 5^k$ and $p$ is a unit vector parallel to $A C$ , then $p$ is equal to

A

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

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B

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

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C

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

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D

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

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E

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

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Solution

The correct option is C $\frac{1}{7} (3^i + 6^j + 2^k)$
Given, $A B =^i + 3^j + 7^k, A D = 2^i + 3^j - 5^k$
$A C = A B + A D$
$= (^i + 3^j + 7^k) + (2^i + 3^j - 5^k)$
$= 3^i + 6^j + 2^k$
Now, $p$ is a unit vector parallel to $A C$ .
Then, $p = \frac{A C}{| A C |}$
$= \frac{3^i + 6^j + 2^k}{\sqrt{9 + 36 + 4}}$
$\Rightarrow p = \frac{1}{7} (3^i + 6^j + 2^k)$

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Similar questions

\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)

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. Find the torque due to three forces (in

Nm

^i + 3^j - 2^k

4^i + 2^j - 3^k

-^i +^j - 5^k

acting on a particle at a point

P (3, 1, 2)

about the point

A (2, - 1, 0)

.

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

⊥

\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)

Let $\to a = \frac{1}{7} (2^i + 3^j + 6^k), \to b = \frac{1}{7} (6^i + 2^j - 3^k), \to c = c_{1}^i + c_{2}^j + c - 3^k$ and matrix $A = ⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{2}{7} & \frac{3}{7} & \frac{6}{7} \frac{6}{7} & \frac{2}{7} & - \frac{3}{7} c_{1} & c_{2} & c_{3} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦$
If $A A^{T} = I, t h e n \to c$ is equal to

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