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The line vecr= veca + lambda vecb will n...

The line `vecr= veca + lambda vecb` will not meet the plane
`vecr cdot vecn =q,` if a.`vecb.vecn=0,veca.vecn=q` b.`vecb.vecn" "ne0,veca.vecn" "ne q` c.`vecb.vecn=0,veca.vecn" "ne q` d.`vecb.vecn" "ne0,veca.vecn" "ne q`

A

`vecb.vecn=0,veca.vecn=q`

B

`vecb.vecn" "ne0,veca.vecn" "ne q`

C

`vecb.vecn=0,veca.vecn" "ne q`

D

`vecb.vecn" "ne0,veca.vecn" "ne q`

Text Solution

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The correct Answer is:
To determine when the line given by the equation \(\vec{r} = \vec{a} + \lambda \vec{b}\) does not meet the plane defined by the equation \(\vec{r} \cdot \vec{n} = q\), we need to analyze the conditions under which the line is parallel to the plane and does not intersect it. ### Step-by-Step Solution: 1. **Understanding the Line and Plane**: - The line is represented as \(\vec{r} = \vec{a} + \lambda \vec{b}\), where \(\vec{a}\) is a point on the line and \(\vec{b}\) is the direction vector of the line. - The plane is given by the equation \(\vec{r} \cdot \vec{n} = q\), where \(\vec{n}\) is the normal vector to the plane. 2. **Condition for Parallelism**: - For the line to be parallel to the plane, the direction vector of the line \(\vec{b}\) must be perpendicular to the normal vector of the plane \(\vec{n}\). This can be expressed mathematically as: \[ \vec{b} \cdot \vec{n} = 0 \] - If this condition holds, the line does not intersect the plane. 3. **Condition for Not Passing Through the Plane**: - Even if the line is parallel to the plane, it must also not pass through any point of the plane. This means that the point \(\vec{a}\) (which lies on the line) should not satisfy the plane's equation: \[ \vec{a} \cdot \vec{n} \neq q \] 4. **Combining the Conditions**: - Therefore, the line \(\vec{r} = \vec{a} + \lambda \vec{b}\) will not meet the plane if: - \(\vec{b} \cdot \vec{n} = 0\) (line is parallel to the plane) - \(\vec{a} \cdot \vec{n} \neq q\) (line does not pass through the plane) 5. **Identifying the Correct Option**: - Based on the conditions derived: - **Option A**: \(\vec{b} \cdot \vec{n} = 0\), \(\vec{a} \cdot \vec{n} = q\) (does not satisfy the second condition) - **Option B**: \(\vec{b} \cdot \vec{n} \neq 0\), \(\vec{a} \cdot \vec{n} \neq q\) (does not satisfy the first condition) - **Option C**: \(\vec{b} \cdot \vec{n} = 0\), \(\vec{a} \cdot \vec{n} \neq q\) (satisfies both conditions) - **Option D**: \(\vec{b} \cdot \vec{n} \neq 0\), \(\vec{a} \cdot \vec{n} \neq q\) (does not satisfy the first condition) - The correct option is **C**: \(\vec{b} \cdot \vec{n} = 0\) and \(\vec{a} \cdot \vec{n} \neq q\).

To determine when the line given by the equation \(\vec{r} = \vec{a} + \lambda \vec{b}\) does not meet the plane defined by the equation \(\vec{r} \cdot \vec{n} = q\), we need to analyze the conditions under which the line is parallel to the plane and does not intersect it. ### Step-by-Step Solution: 1. **Understanding the Line and Plane**: - The line is represented as \(\vec{r} = \vec{a} + \lambda \vec{b}\), where \(\vec{a}\) is a point on the line and \(\vec{b}\) is the direction vector of the line. - The plane is given by the equation \(\vec{r} \cdot \vec{n} = q\), where \(\vec{n}\) is the normal vector to the plane. ...
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CENGAGE ENGLISH-THREE-DIMENSIONAL GEOMETRY -SINGLE CORRECT ANSWER TYPE
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