MCQ
Three points whose position vectors are $a + b,\,\,a - b$ and $a + kb$ will be collinear, if the value of $k $ is
- AZero
- BOnly negative real number
- COnly positive real number
- ✓Every real number
Here $\overrightarrow {AB} = - 2b,$ $\overrightarrow {BC} = (k + 1)b$
Hence $\forall \,\,k \in R \Rightarrow \overrightarrow {AB} = \lambda \overrightarrow {BC} .$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$STATEMENT -1$ : The tangents are mutually perpendicular. because
$STATEMENT - 2$ : The locus of the points from which mutually perpendicular tangents can be drawn to the given circle is $x^2+y^2=338$
$(A)$ $P(E)=\frac{4}{5}, P(F)=\frac{3}{5}$
$(B)$ $P(E)=\frac{1}{5}, P(F)=\frac{2}{5}$
$(C)$ $P(E)=\frac{2}{5}, P(F)=\frac{1}{5}$
$(D)$ $P(E)=\frac{3}{5}, P(F)=\frac{4}{5}$