When calculating direction cosines for a line perpendicular to two given lines, always start by finding the cross product of the direction ratios of the lines. Once you have the resulting vector, normalize it by dividing each component by the magnitude of the vector. This will give you the direction cosines. Remember that the cross product is a fundamental operation when dealing with perpendicular vectors in 3D space.
To find the direction cosines of a line perpendicular to the given lines, we need to find a vector that is perpendicular to both vectors represented by the direction ratios (1, -2, -2) and (0, 2, 1). We achieve this by computing the cross product of these two vectors:
\( \mathbf{a} = (1, -2, -2) \)
\( \mathbf{b} = (0, 2, 1) \)
The cross product \( \mathbf{a} \times \mathbf{b} \) is given by:
\(\mathbf{c} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -2 & -2 \\ 0 & 2 & 1 \end{vmatrix}\)
\(\mathbf{c} = \mathbf{i}((-2)(1) - (-2)(2)) - \mathbf{j}((1)(1) - (-2)(0)) + \mathbf{k}((1)(2) - (-2)(0))\)
\(\mathbf{c} = \mathbf{i}(-2 + 4) - \mathbf{j}(1) + \mathbf{k}(2)\)
\(\mathbf{c} = 2\mathbf{i} - \mathbf{j} + 2\mathbf{k}\)
Thus, the direction ratios of the perpendicular line are (2, -1, 2).
Now, we convert these direction ratios to direction cosines by dividing each value by the magnitude of the vector.
Magnitude of vector: \(\sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3\)
So, the direction cosines are:
\(\frac{2}{3}, -\frac{1}{3}, \frac{2}{3}\)
Thus, the correct option is \( \frac{2}{3}, -\frac{1}{3}, \frac{2}{3} \).
Use the concept that the direction cosines of a line perpendicular to two given lines can be found by taking the cross product of the direction ratios.
We are given the direction ratios of two lines: \( \langle 1, -2, -2 \rangle \) and \( \langle 0, 2, 1 \rangle \). To find the direction cosines of the line perpendicular to these two lines, we compute the cross product of these two vectors.
Step 1: Compute the cross product of the vectors:
The cross product of two vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \) is given by:
\[ \mathbf{a} \times \mathbf{b} = \langle a_2 b_3 - a_3 b_2, a_3 b_1 - a_1 b_3, a_1 b_2 - a_2 b_1 \rangle \]
Substituting the values \( \langle 1, -2, -2 \rangle \) for \( \mathbf{a} \) and \( \langle 0, 2, 1 \rangle \) for \( \mathbf{b} \), we get:
\[ \mathbf{a} \times \mathbf{b} = \langle (-2)(1) - (-2)(2), (-2)(0) - (1)(1), (1)(2) - (-2)(0) \rangle \]
Simplifying the components:
\[ \mathbf{a} \times \mathbf{b} = \langle -2 + 4, 0 - 1, 2 - 0 \rangle = \langle 2, -1, 2 \rangle \]
Step 2: Normalize the resulting vector to get the direction cosines:
The direction cosines are obtained by normalizing the vector \( \langle 2, -1, 2 \rangle \). To normalize, we divide each component of the vector by its magnitude.
The magnitude of \( \langle 2, -1, 2 \rangle \) is:
\[ |\mathbf{v}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \]
Now, divide each component by the magnitude:
\[ \text{Direction cosines} = \left\langle \frac{2}{3}, \frac{-1}{3}, \frac{2}{3} \right\rangle \]
Conclusion:
The direction cosines of the line perpendicular to the two given lines are \( \left\langle \frac{2}{3}, \frac{-1}{3}, \frac{2}{3} \right\rangle \). This confirms that option (1) is the correct answer.
Let a line passing through the point $ (4,1,0) $ intersect the line $ L_1: \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} $ at the point $ A(\alpha, \beta, \gamma) $ and the line $ L_2: x - 6 = y = -z + 4 $ at the point $ B(a, b, c) $. Then $ \begin{vmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{vmatrix} \text{ is equal to} $
Rearrange the parts to form a coherent sentence:
A) when it is no longer fun.
B) stop doing something
C) if you're not growing
D) or learning from it
A consumer experiences the following total utility from consuming a certain good:
If the price per unit is ₹4, at what quantity does the consumer stop purchasing under the equilibrium condition where M U m = 5?
The Darsanams of the Gosangi
Over the costumes, Gosangi wears various objects made up of leather, shells, metal and threads as ornaments. Traditionally, the prominent among them is known as Darsanam-s, which literally means vision or suggesting that which is visible. There are altogether seven Darsanams, which can be neither considered as costumes nor ornaments. But, for an outsider, they may look like ornaments. The first Darsanam that Gosangi wears,cover chest and the back. This is traditionally identified as Rommu Darsanam or Sanku Darsanam. The second one is tied around the neck and called as Kanta Darsanam. The third and fourth ones are tied around the arms of left and right hands. The fifth and sixth ones are tied to the left and right wrists. (For these specific names are mentioned by the performers). The seventh one is known as Siro Darsanam, and it is tied around the already tied hair (koppu). The performers also know all these Dasanam except the Rommu Darsanam and Dasthavejulu (records).
Percussive Musical Instruments of India
India is very rich in the number and variety of musical instruments. From time immemorial, musical instruments have been connected with various Gods and goddesses according to mythol ogy. Musical Instruments have been classified into Thata, Avanadha, Ghana and Sushira. We came across this classification first in Natyashastra. Thata variety, is an instrument with strings and played by plucking or bowing. The instruments like Veena, Sitar, violin, Sarangi etc. come under this category. The Avandha variety are instruments with skin-covered heads, and are played by beating on both sides or one side. Mridangam, Pakhawaj, Tabla etc. come under this category. Ghana vadyas are those made with metal content. Manjira, Ghatom etc. are some of the examples of Ghana Vadya. Sushira Vadya are those instruments with holes and make the sound by blowing air through the holes. Flute, Nagaswaram, Saxophone, Clarinet are some of the examples.