Three Dimensional Geometry — Class 12 Maths Solution

ncert exercise SA NCERT,EX.11.2,Q.7, Page .477
Question

The cartesian equation of a line is $\cfrac{{x - 5}}{3} = \cfrac{{y + 4}}{5} = \cfrac{{z + 5}}{6}$ .
Write its vector form.

Step-by-step Solution

The given cartesian equation (in symmetrical form) is
$\cfrac{{x - 5}}{3} = \cfrac{{y - ( - 4)}}{7} = \cfrac{{z - 6}}{2}$

..(1)
$\Rightarrow$ The line passes through the point $(5, - 4,6)$ and is parallel to vector $3\hat i + 7\hat j + 2\hat k.$

Hence, the vector equation of the line is
$\vec r = (5\hat i - 4\hat j + 6\hat k) + \lambda (3\hat i + 7\hat j + 2\hat k)$ .

NCERT & Exemplar solution for CBSE Class 12 Mathematics, Three Dimensional Geometry. Curated by Sachin Sharma. Free for all students.