Let the shortest distance between the line $ L: \frac{x-5}{-2}=\frac{y-\lambda}{1}=\frac{z+\lambda}{1}, \lambda \geq 0 $ and $ L_{1}: x+1=y- $ $ 1=4-z $ be $ 2 \sqrt{6} $. If $ (\alpha, \beta, \gamma) $ lies on $ L $, then which of the following is NOT possible?

JEE MATHEMATICS MCQ: SCALAR TRIPLE PRODUCT

1 Easy

1 of 1

Analyzing +4 | -1

Let the shortest distance between the line

L: \frac{x-5}{-2}=\frac{y-\lambda}{1}=\frac{z+\lambda}{1}, \lambda \geq 0

and

L_{1}: x+1=y-

1=4-z

2 \sqrt{6}

. If

(\alpha, \beta, \gamma)

lies on

L

, then which of the following is NOT possible?

View answer and explanation

Correct option: A

\alpha+2 \gamma=24

Option A is the correct answer because it is the only option that is not possible given the constraints of the problem and the calculated value of lambda.

More Questions from SCALAR TRIPLE PRODUCT

Practice All Chapter MCQs →

\vec r\cdot\vec a = \vec r\cdot\vec b = \vec r\cdot\vec c = 0

for some non-zero vector

\vec r,

then the value of

\left[ {\vec a\vec b\vec c} \right],

is The shortest distance between the lines

\frac{{x - 3}}{2} = \frac{{y - 2}}{3} = \frac{{z - 1}}{{ - 1}}

and

\frac{{x + 3}}{2} = \frac{{y - 6}}{1} = \frac{{z - 5}}{3},

is Let

\mathbf{\vec a} = \widehat \mathbf{i} - \mathbf{\hat k},\;\mathbf{\vec b} = x\widehat \mathbf{i} + \widehat \mathbf{j} + \left( {1 - x} \right)\mathbf{\hat k}

and

\mathbf{\vec c} = y\widehat \mathbf{i} + x\widehat \mathbf{j} + \left( {1 + x - y} \right)\mathbf{\hat k}.

Then

\left[ {\mathbf{\vec a},\;\mathbf{\vec b},\;\mathbf{\vec c}} \right]

depends on If

\vec{a}, \vec{b}

are non-zero vectors, then

((\vec{a} \times \vec{b}) \times \vec{a}) \cdot((\vec{b} \times \vec{a}) \times \vec{b})

equals

a \cdot(a \times b)=

The length of the perpendicular from the origin to the plane passing through the point a and containing the line

\mathbf{r}=\mathbf{b}+\lambda \mathbf{c}