Let overset arrow x and overset arrow y are 2 non-zero and non-collinear vectors, then the largest value of k such that the non-zero vectors (k2 - 5 k + 6) overset arrow x+(k - 3) overset arrow y and 2 overset arrow x+5 overset arrow y are collinear is

Question

Let overset arrow x and overset arrow y are 2 non-zero and non-collinear vectors, then the largest value of k such that the non-zero vectors (k2 - 5 k + 6) overset arrow x+(k - 3) overset arrow y and 2 overset arrow x+5 overset arrow y are collinear is (A) 3 (B) 6 (C) (12/5) (D) -1

Tardigrade Admin · Accepted Answer

Let, overset arrow a=(k2 - 5 k + 6) overset arrow x+(k - 3) overset arrow y and overset arrow b=2 overset arrow x+5 overset arrow y ∵ overset arrow a and overset arrow b are collinear vectors ⇒ (k2 - 5 k + 6/2)=(k - 3/5) ⇒ k=3,(12/5) But, for k=3, overset arrow a=0 (not possible)

Q. Let $x \to$ and $y \to$ are $2$ non-zero and non-collinear vectors, then the largest value of $k$ such that the non-zero vectors $(k^{2} - 5 k + 6) x \to + (k - 3) y \to$ and $2 x \to + 5 y \to$ are collinear is

$3$

$6$

$\frac{12}{5}$

$- 1$

Let, $a \to = (k^{2} - 5 k + 6) x \to + (k - 3) y \to$
and $b \to = 2 x \to + 5 y \to$
$∵ a \to$ and $b \to$ are collinear vectors
$\Rightarrow \frac{k ^{2} - 5 k + 6}{2} = \frac{k - 3}{5}$
$\Rightarrow k = 3, \frac{12}{5}$
But, for $k = 3, a \to = 0$ (not possible)

Q. Let x→ and y→​ are 2 non-zero and non-collinear vectors, then the largest value of k such that the non-zero vectors (k2−5k+6)x→+(k−3)y→​ and 2x→+5y→​ are collinear is

3

6

512​

−1

Let, a→=(k2−5k+6)x→+(k−3)y→​ and b→=2x→+5y→​ ∵a→ and b→ are collinear vectors ⇒2k2−5k+6​=5k−3​ ⇒k=3,512​ But, for k=3,a→=0 (not possible)

Q. Let $x \to$ and $y \to$ are $2$ non-zero and non-collinear vectors, then the largest value of $k$ such that the non-zero vectors $(k^{2} - 5 k + 6) x \to + (k - 3) y \to$ and $2 x \to + 5 y \to$ are collinear is

$3$

$6$

$\frac{12}{5}$

$- 1$

Let, $a \to = (k^{2} - 5 k + 6) x \to + (k - 3) y \to$
and $b \to = 2 x \to + 5 y \to$
$∵ a \to$ and $b \to$ are collinear vectors
$\Rightarrow \frac{k ^{2} - 5 k + 6}{2} = \frac{k - 3}{5}$
$\Rightarrow k = 3, \frac{12}{5}$
But, for $k = 3, a \to = 0$ (not possible)