Let AB be the line through the points (4,7,8), and (2,3,4), and CD be the line through the points (-1,-2,1), and (1,2,5).
The direction ratios, a1, b1, c1 of AC are (2-4), (3-7), and (4-8) i.e., -2, -4, and -4.
The direction ratios, a2, b2, c2 of CD are (1-(-1)), (2-(-2)), and (5-1) i.e., 2, 4, and 4.
AB will parellel to CD, if \(\frac{a_1}{a_2}\)=\(\frac{b_1}{b_2}\)=\(\frac{c_1}{c_2}\)
\(\frac{a_1}{a_2}\)
=\(-\frac{2}{2}\)
=-1
\(\frac{b_1}{b_2}\)
=\(-\frac{4}{4}\)
=-1
\(\frac{c_1}{c_2}\)
=\(-\frac{4}{4}\)
=-1
∴ \(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\) = \(\frac{c_1}{c_2}\)
Thus, AB is parallel to CD.
List - I | List - II | ||
(P) | γ equals | (1) | \(-\hat{i}-\hat{j}+\hat{k}\) |
(Q) | A possible choice for \(\hat{n}\) is | (2) | \(\sqrt{\frac{3}{2}}\) |
(R) | \(\overrightarrow{OR_1}\) equals | (3) | 1 |
(S) | A possible value of \(\overrightarrow{OR_1}.\hat{n}\) is | (4) | \(\frac{1}{\sqrt6}\hat{i}-\frac{2}{\sqrt6}\hat{j}+\frac{1}{\sqrt6}\hat{k}\) |
(5) | \(\sqrt{\frac{2}{3}}\) |
Rupal, Shanu and Trisha were partners in a firm sharing profits and losses in the ratio of 4:3:1. Their Balance Sheet as at 31st March, 2024 was as follows:
(i) Trisha's share of profit was entirely taken by Shanu.
(ii) Fixed assets were found to be undervalued by Rs 2,40,000.
(iii) Stock was revalued at Rs 2,00,000.
(iv) Goodwill of the firm was valued at Rs 8,00,000 on Trisha's retirement.
(v) The total capital of the new firm was fixed at Rs 16,00,000 which was adjusted according to the new profit sharing ratio of the partners. For this necessary cash was paid off or brought in by the partners as the case may be.
Prepare Revaluation Account and Partners' Capital Accounts.
In a plane, the equation of a line is given by the popular equation y = m x + C. Let's look at how the equation of a line is written in vector form and Cartesian form.
Consider a line that passes through a given point, say ‘A’, and the line is parallel to a given vector '\(\vec{b}\)‘. Here, the line ’l' is given to pass through ‘A’, whose position vector is given by '\(\vec{a}\)‘. Now, consider another arbitrary point ’P' on the given line, where the position vector of 'P' is given by '\(\vec{r}\)'.
\(\vec{AP}\)=𝜆\(\vec{b}\)
Also, we can write vector AP in the following manner:
\(\vec{AP}\)=\(\vec{OP}\)–\(\vec{OA}\)
𝜆\(\vec{b}\) =\(\vec{r}\)–\(\vec{a}\)
\(\vec{a}\)=\(\vec{a}\)+𝜆\(\vec{b}\)
\(\vec{b}\)=𝑏1\(\hat{i}\)+𝑏2\(\hat{j}\) +𝑏3\(\hat{k}\)