The set of all \alpha , for which the vectors \veca=\alpha t \hati+6 \hatj-3 \hatk and \vecb=t \hati-2 \hatj-2 \alpha t \hatk are inclined at an obtuse angle for all t \in \mathbbR , is

Vector Algebra PYQ — JEE Main 2024 · 8 Apr · Shift 1 · Q34 with Solution

View written solutionFree

Correct answer: B

For two vectors to be inclined at an obtuse angle, their dot product must be negative:

\vec a \cdot \vec b < 0

for all $t \in \mathbb R$ .

Here,

\vec a = \alpha t\,\hat i + 6\,\hat j - 3\,\hat k, \qquad \vec b = t\,\hat i - 2\,\hat j - 2\alpha t\,\hat k

Compute the dot product:

\vec a \cdot \vec b = (\alpha t)(t) + (6)(-2) + (-3)(-2\alpha t)

= \alpha t^2 - 12 + 6\alpha t

So we need

\alpha t^2 + 6\alpha t - 12 < 0 \quad \text{for all } t \in \mathbb R

f(t)=\alpha t^2 + 6\alpha t - 12

We want $f(t)<0$ for all real $t$ .

This is a quadratic in $t$ .

For a quadratic to be negative for all real $t$ , the following are needed:

coefficient of $t^2$ is negative: $\alpha<0$
discriminant is non-positive: $D\le 0$

Compute the discriminant:

D = (6\alpha)^2 - 4(\alpha)(-12)

= 36\alpha^2 + 48\alpha

= 12\alpha(3\alpha+4)

Now require

12\alpha(3\alpha+4) \le 0

which gives

\alpha \in \left[-\frac43,0\right]

But we also need $\alpha<0$ if it is genuinely quadratic and negative for all $t$ .

Check the endpoint $\alpha=0$ separately.

If $\alpha=0$ , then

f(t)=-12<0 \quad \text{for all } t

So $\alpha=0$ is allowed.

Check the endpoint $\alpha=-\frac43$ :

Then discriminant is $0$ , so the quadratic touches zero at one value of $t$ . Indeed,

f(t)= -\frac43 t^2 - 8t - 12

At its vertex it becomes $0$ , so the angle is a right angle there, not obtuse. Hence $\alpha=-\frac43$ is not allowed.

Therefore,

\alpha \in \left(-\frac43,0\right]

Compare with options:

Option B is

\left(-\frac43,0\right]

which matches exactly.

More from Vector Algebra

View written solutionFree

Correct answer: B

For two vectors to be inclined at an obtuse angle, their dot product must be negative:

\vec a \cdot \vec b < 0

for all $t \in \mathbb R$ .

Here,

\vec a = \alpha t\,\hat i + 6\,\hat j - 3\,\hat k, \qquad \vec b = t\,\hat i - 2\,\hat j - 2\alpha t\,\hat k

Compute the dot product:

\vec a \cdot \vec b = (\alpha t)(t) + (6)(-2) + (-3)(-2\alpha t)

= \alpha t^2 - 12 + 6\alpha t

So we need

\alpha t^2 + 6\alpha t - 12 < 0 \quad \text{for all } t \in \mathbb R

f(t)=\alpha t^2 + 6\alpha t - 12

We want $f(t)<0$ for all real $t$ .

This is a quadratic in $t$ .

For a quadratic to be negative for all real $t$ , the following are needed:

coefficient of $t^2$ is negative: $\alpha<0$
discriminant is non-positive: $D\le 0$

Compute the discriminant:

D = (6\alpha)^2 - 4(\alpha)(-12)

= 36\alpha^2 + 48\alpha

= 12\alpha(3\alpha+4)

Now require

12\alpha(3\alpha+4) \le 0

which gives

\alpha \in \left[-\frac43,0\right]

But we also need $\alpha<0$ if it is genuinely quadratic and negative for all $t$ .

Check the endpoint $\alpha=0$ separately.

If $\alpha=0$ , then

f(t)=-12<0 \quad \text{for all } t

So $\alpha=0$ is allowed.

Check the endpoint $\alpha=-\frac43$ :

Then discriminant is $0$ , so the quadratic touches zero at one value of $t$ . Indeed,

f(t)= -\frac43 t^2 - 8t - 12

At its vertex it becomes $0$ , so the angle is a right angle there, not obtuse. Hence $\alpha=-\frac43$ is not allowed.

Therefore,

\alpha \in \left(-\frac43,0\right]

Compare with options:

Option B is

\left(-\frac43,0\right]

which matches exactly.

Vector Algebra question

Vector Algebra question

More from Vector Algebra

Vector Algebra question

Vector Algebra question

More from Vector Algebra